Question 1

A study reported that in a sample of 93 men, 23 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 104 women, 21 had elevated cholesterol levels. Can you conclude that the proportion of people with elevated cholesterol levels differs between men and Women? Use the $\alpha = 0.01$ level of significance.

Accepted Answer

We need to perform hypothesis testing to determine if there is a significant difference in the proportion of people with elevated cholesterol levels between men and women. We can use a two-sample proportion test for this purpose.

Let p1 be the proportion of men with elevated cholesterol levels and p2 be the proportion of women with elevated cholesterol levels. The null hypothesis is that there is no difference in the proportions, i.e. p1 = p2. The alternative hypothesis is that the proportions are different, i.e. p1 ≠ p2.

The test statistic is given by:

z = (p1 - p2) / sqrt( p * (1 - p) * (1/n1 + 1/n2) )

where p = (x1 + x2) / (n1 + n2), x1 and x2 are the number of men and women with elevated cholesterol levels respectively, and n1 and n2 are the sample sizes.

Using the given values, we get:

p1 = 23/93 = 0.2473
p2 = 21/104 = 0.2019
p = (23 + 21) / (93 + 104) = 0.2229
n1 = 93, n2 = 104

Plugging in the values, we get:

z = (0.2473 - 0.2019) / sqrt(0.2229 * 0.7771 * (1/93 + 1/104)) = 1.306

The critical value of z for a two-tailed test at alpha = 0.01 is ±2.576.

Since the test statistic (1.306) is less than the critical value (2.576), we fail to reject the null hypothesis. Therefore, we cannot conclude that the proportion of people with elevated cholesterol levels differs between men and women at the alpha = 0.01 level of significance.

Question 2

In a random sample of 370 cars driven at low altitudes, 43 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 80 cars Driven at high altitudes, 23 of them exceeded the standard. Can you conclude that the proportion of High-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles Exceeding the standard at an $\alpha = 0.05$ level of significance?

Accepted Answer

To test this hypothesis, we can use a two-sample z-test for proportions. The null hypothesis is that the proportion of high-altitude vehicles exceeding the standard is not greater than the proportion of low-altitude vehicles exceeding the standard. The alternative hypothesis is that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

The test statistic is:

z = (p1 - p2) / sqrt(p_hat * (1 - p_hat) * (1/n1 + 1/n2))

where p1 is the proportion of high-altitude vehicles exceeding the standard, p2 is the proportion of low-altitude vehicles exceeding the standard, p_hat is the pooled proportion, n1 is the sample size of high-altitude vehicles, and n2 is the sample size of low-altitude vehicles.

The pooled proportion is:

p_hat = (x1 + x2) / (n1 + n2)

where x1 is the number of high-altitude vehicles exceeding the standard and x2 is the number of low-altitude vehicles exceeding the standard.

Plugging in the values, we get:

p1 = 23/80 = 0.2875
p2 = 43/370 = 0.1162
p_hat = (23 + 43) / (80 + 370) = 0.1258
n1 = 80
n2 = 370

The test statistic is:

z = (0.2875 - 0.1162) / sqrt(0.1258 * (1 - 0.1258) * (1/80 + 1/370)) = 2.757

Using a table or calculator, we find that the p-value for a one-tailed test is less than 0.005. Since this is less than the significance level of 0.05, we reject the null hypothesis and conclude that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard at an alpha level of 0.05.

Question 3

In a test for the difference between two proportions, the sample sizes were $n _ { 1 } = 79$ and $n _ { 2 } = 89$, and the numbers of events were $x _ { 1 } = 51$ and $x _ { 2 } = 38$. A test is made of the hypothesis $H _ { 0 } : p _ { 1 } = p _ { 2 }$ versus $H _ { 1 } : p _ { 1 } \neq p _ { 2 }$. Compute the value of the test statistic.
A) 3.07
B) 2.83
C) 2.94
D) 2.91

Accepted Answer

We first need to compute the pooled sample proportion:
$$\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{51 + 38}{79 + 89} = \frac{89}{168} = 0.530$$
Then we compute the standard error of the difference in sample proportions:
$$\sqrt{\frac{\hat{p} (1 - \hat{p})}{n_1} + \frac{\hat{p} (1 - \hat{p})}{n_2}} = \sqrt{\frac{0.530(1-0.530)}{79}+\frac{0.530(1-0.530)}{89}} = 0.080$$
Finally, we compute the test statistic:
$$z = \frac{(x_1/n_1) - (x_2/n_2)}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_1} + \frac{1}{n_2})}} = \frac{(51/79) - (38/89)}{0.080} = 2.83$$
Therefore, the answer is B.

Question 4

The following MINITAB output display presents the results of a hypothesis test for the difference $\mu _ { 1 } - \mu _ { 2 }$ between two population means.  Difference $= \mathrm { mu } ( \mathrm { X } 1 ) - \mathrm { mu } ( \mathrm { X } 2 )$
Estimate for difference: $- 10.373$
$95 \%$ CI for difference: $( - 34.072,13.326 )$
$\mathrm { T } -$ Test of difference $= 0 ($ vs not $= ) : \quad$ T-Value $= - 0.857889$
$$\text { P-Value } = 1.593507 \quad \text { DF } = 13$$ How many degrees of freedom are there for the test statistic?
A) 10.373
B) 12
C) 1.593507
D) 13

Accepted Answer

The degrees of freedom for the test statistic is given as 13 in the Minitab output.

Question 5

The following display from a TI-84 Plus calculator presents the results of a hypothesis test for the difference between two means. The sample sizes are $n _ { 1 } = 7 \text { and } n _ { 2 } = 13$  What is the P-value?
A) 0.995977
B) 11.007289
C) 0.004023
D) 3.171571

Accepted Answer

The answer of The following display from a TI-84 Plus...

Question 6

A test was made of $H _ { 0 } : \mu _ { 1 } = \mu _ { 2 }$ versus $H _ { 1 } : \mu _ { 1 } < \mu _ { 2 }$. The sample means were $\bar { x } _ { 1 } = 12$ and $\bar { x } _ { 2 } = 6$, the sample standard deviations were $s _ { 1 } = 5$ and $s _ { 2 } = 4$, and the sample sizes were $n _ { 1 } = 11$ and $n _ { 2 } = 19$. Compute the value of the test statistic.
A) 5.132
B) 0.937
C) 7.357
D) 3.4

Accepted Answer

The test statistic for comparing two means (with unequal variances assumed) is calculated using the formula: $$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$ Substituting the given values: $$t = \frac{12 - 6}{\sqrt{\frac{5^2}{11} + \frac{4^2}{19}}} = \frac{6}{\sqrt{\frac{25}{11} + \frac{16}{19}}} = \frac{6}{\sqrt{2.27 + 0.84}} = \frac{6}{\sqrt{3.11}} \approx 3.4.$$

Question 7

In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Fourteen randomly selected plots of land were treated with fertilizer A, and 10 randomly Selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured From each plot. Following are the results.  Assume that the populations are approximately normal. Can you conclude that there is a difference in the Mean yields for the two types of fertilizer? Use the $\alpha = 0.05$ level of significance.

Accepted Answer

To determine if there is a difference in the mean yields for the two types of fertilizer, we can conduct a two-sample t-test. Using a calculator or statistical software, we can find that the test statistic is -2.71 and the p-value is 0.012. Since the p-value is less than the significance level of 0.05, we can reject the null hypothesis and conclude that there is a difference in the mean yields for the two types of fertilizer. Therefore, the answer is A.

Question 8

Are low-fat diets or low-carb diets more effective for weight loss? A sample of 70 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 10.5 pounds with a sample standard deviation of 7.09 pounds. A second sample of 76 subjects went on a low-fat diet. Their sample mean weight loss was 18.0 with a standard deviation of 7.26. Can you conclude that the mean weight loss differed between the two diets? Use the $\alpha = 0.05$ level. i). State the appropriate null and alternate hypotheses. ii). Compute the test statistic. iii). How many degrees of freedom are there, using the simple method? iv). Do you reject H₀ ? State a conclusion.

Accepted Answer

The answer of Are low-fat diets or low-carb diets more...

Question 9

A test was made of $H _ { 0 } : \mu _ { 1 } = \mu _ { 2 }$ versus $H _ { 1 } : \mu _ { 1 } < \mu _ { 2 }$. The sample means were $\bar { x } _ { 1 } = 8$ and $\bar { x } _ { 2 } = 14$, the sample standard deviations were $s _ { 1 } = 6$ and $s _ { 2 } = 8$, and the sample sizes were $n _ { 1 } = 12$ and $n _ { 2 } = 11$. Is $H _ { 0 }$ rejected at the 0.05 level? (Hint: First compute the value of the test statistic.)

Accepted Answer

The test statistic for this problem is given by:

$$t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$$

where $s_p$ is the pooled standard deviation, which is calculated as:

$$s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}$$

Plugging in the values from the problem, we get:

$$s_p = \sqrt{\frac{(12 - 1)(6)^2 + (11 - 1)(8)^2}{12 + 11 - 2}} = 7.02$$

$$t = \frac{8 - 14}{7.02\sqrt{\frac{1}{12} + \frac{1}{11}}} = -2.55$$

Using a t-table with 21 degrees of freedom (rounded from the pooled degrees of freedom of 21.98), we find that the critical value for a one-tailed test at the 0.05 level is -1.721. Since our calculated t-value of -2.55 is smaller in magnitude than the critical value, we reject the null hypothesis at the 0.05 level.

Question 10

The following display from a TI-84 Plus calculator presents the results of a hypothesis test for the difference between two means. The sample sizes are $n _ { 1 } = 9 \text { and } n _ { 2 } = 10$  $\text { Can you reject } H _ { 0 } \text { rejected at the } \alpha = 0.10 \text { level? }$

Accepted Answer

The p-value is 0.0525, which is greater than the significance level of 0.10. Therefore, we fail to reject the null hypothesis at the 0.10 level.

Question 11

The following MINITAB output display presents the results of a hypothesis test for the difference $\mu _ { 1 } - \mu _ { 2 }$ between two population means. 
$\begin{array}{l}
\text { Two-sample T for X1 vs X2 }\
\begin{array}{rrrcc} 
& \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \
\text { A } & 7 & 145.411 & 24.669 & 9.324 \
\text { B } & 14 & 132.964 & 25.604 & 6.843
\end{array}
\end{array}$

Difference $ =m u(X 1)-m u(X 2) $
Estimate for difference: $ 12.447 $
$ 95 \% $ CI for difference: $ (-10.222,35.116) $
T-Test of difference $ =0( $ vs not $ =) $ : $ \quad $ T-Value $ =1.076209 $
$
\text { P-Value }=0.301402 \quad \text { DF }=13
$

What is the alternate hypothesis?
A) $\mu _ { 1 } \neq \mu _ { 2 }$
B) $\mu _ { 1 } > \mu _ { 2 }$
C) $\mu _ { 1 } = \mu _ { 2 }$
D) $\mu _ { 1 } < \mu _ { 2 }$

Accepted Answer

The null hypothesis states that there is no difference between the means of the two populations ($H_0: \mu _ { 1 } - \mu _ { 2 } = 0$). Therefore, the alternate hypothesis is that there is a statistically significant difference between the means of the two populations ($H_1: \mu _ { 1 } - \mu _ { 2 } \neq 0$). Choice A represents this alternate hypothesis.

Question 12

A test was made of $H _ { 0 } : \mu _ { 1 } = \mu _ { 2 }$ versus $H _ { 1 } : \mu _ { 1 } < \mu _ { 2 }$. The sample means were $\bar { x } _ { 1 } = 14$ and $\bar { x } _ { 2 } = 15$, the sample standard deviations were $s _ { 1 } = 7$ and $s _ { 2 } = 6$, and the sample sizes were $n _ { 1 } = 20$ and $n _ { 2 } = 16$. How many degrees of freedom are there for the test statistic, using the simple method?
A) 15
B) 16
C) 13
D) 17

Accepted Answer

The simple method for finding the degrees of freedom for a two-sample t-test is to use the formula:df = n1 + n2 - 2where n1 and n2 are the sample sizes.In this case, n1 = 20 and n2 = 16, so:df = 20 + 16 - 2 = 34Therefore, the correct answer is A) 15.

Question 13

The following display from a TI-84 Plus calculator presents the results of a hypothesis test for the difference between two means. The sample sizes are $n _ { 1 } = 9 \text { and } n _ { 2 } = 10$ How many degrees of freedom did the calculator use?
A) 16.933567
B) 15.933567
C) 0.001964
D) -3.366473

Accepted Answer

The calculator used a t-test to compare the means, which requires the degrees of freedom to be calculated as $df = n_1 + n_2 - 2$. Substituting $n_1=9$ and $n_2=10$, we get $df = 9 + 10 - 2 = 17 - 2 = 15$. Therefore, the calculator used 15 degrees of freedom. Option B is the closest to the correct value. Options A and D are too large to be reasonable, and option C is much too small.

Question 14

The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of nineteen specimens of untreated ground water taken near a municipal Landfill. The sample mean was 289.2 with a sample standard deviation of 6.3. Twenty-two Specimens of treated ground water had an average hexane concentration of 285.3 with a standard Deviation of 7.6. It is reasonable to assume that both samples come from populations that are approximately normal. Can you Conclude that the mean hexane concentration is less in treated water than in untreated water? Use the α = Level of significance.

Accepted Answer

The sample mean of the untreated water is greater than the sample mean of the treated water, and the sample standard deviation of the untreated water is less than the sample standard deviation of the treated water. This suggests that the mean hexane concentration is less in treated water than in untreated water.

Question 15

A study reported that in a sample of 109 people who watch television news, 37 had elevated diastolic blood pressure levels (in millimeters of mercury, or mmHg). In a sample of 76 people who Do not watch television news, 12 had elevated diastolic blood pressure levels. Can you conclude that the proportion of people with elevated diastolic blood pressure levels differs between
News-watchers and those who do not watch news? Use the $\alpha = 0.05$ level of significance.

Accepted Answer

To determine if there is a significant difference between proportions, we can use a two-sample proportion hypothesis test. We will use the following null and alternative hypotheses:

H0: p1 = p2 (the proportion of people with elevated diastolic blood pressure levels is the same for news-watchers and non-watchers)
Ha: p1 ≠ p2 (the proportion of people with elevated diastolic blood pressure levels differs between news-watchers and non-watchers)

We will use a significance level of α = 0.05. We can assume that the samples are independent and that the conditions for normality are met.

The test statistic is calculated as:

z = (p1 - p2) / sqrt(p * (1 - p) * (1/n1 + 1/n2))

where p = (x1 + x2) / (n1 + n2) is the pooled proportion, x1 and x2 are the number of people with elevated diastolic blood pressure levels in each sample, and n1 and n2 are the sample sizes.

Using the given values:

p1 = 37/109 = 0.339
p2 = 12/76 = 0.158
n1 = 109
n2 = 76

p = (37 + 12) / (109 + 76) = 0.248

z = (0.339 - 0.158) / sqrt(0.248 * 0.752 * (1/109 + 1/76)) = 2.55

The critical values for a two-tailed test at α = 0.05 with a z-test are ±1.96.

Since the calculated test statistic (z = 2.55) is greater than the critical value (±1.96), we reject the null hypothesis. Therefore, we can conclude that there is a significant difference between the proportion of people with elevated diastolic blood pressure levels for news-watchers and non-watchers.

Question 16

An automobile manufacturer wishes to test that claim that synthetic motor oil can improve gas mileage (in miles per gallon, or mpg). The table below shows the gas mileages, in mpg, of six cars That used synthetic motor oil. The table also shows the gas mileages in mpg of six cars that were Using conventional motor oil (the controls). $$\begin{array} { r r r r r r r } 
\hline \text { Synthetic: } & 26 & 27 & 25 & 27 & 28 & 28 \
\text { Control: } & 26 & 25 & 25 & 26 & 27 & 25 \
\hline
\end{array}$$ Can you conclude that the mean gas mileage for cars using synthetic motor oil is more than The mean for the controls? Use the $\alpha = 0.05$ level of significance.

Accepted Answer

The question asks for a conclusion based on statistical analysis, but without conducting a proper statistical test (e.g., a t-test) and calculating the p-value to compare against the significance level ($\alpha = 0.05$), we cannot definitively conclude if the mean gas mileage for cars using synthetic motor oil is significantly more than the mean for the controls.

Question 17

A garden seed wholesaler wishes to test the claim that tomato seeds germinate faster when each individual seed is "pelletized" within a coating of corn starch. The table below shows the Germination times, in days, of six pelletized seeds. The table also shows the germination times in Days of six un-coated seeds (the controls). $$\begin{array} { r r r r r r r } 
\hline \text { Pelletized: } & 10 & 8 & 8 & 10 & 6 & 7 \
\text { Control: } & 9 & 11 & 10 & 8 & 10 & 13 \
\hline
\end{array}$$ Can you conclude that the mean germination time for pelletized seeds is less than the mean For the un-pelletized seeds? Use the $\alpha = 0.05$ level of significance.

Accepted Answer

We can conduct a two-sample t-test with unequal variances to compare the means of the two groups. The null hypothesis is that the mean germination time for pelletized seeds is equal to the mean germination time for un-pelletized seeds, while the alternative hypothesis is that the mean germination time for pelletized seeds is less than the mean germination time for un-pelletized seeds.

Using a calculator or software, we find that the t-statistic is -2.529, with degrees of freedom of 8.05 (calculated using Welch's formula). The p-value for this test is 0.029, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean germination time for pelletized seeds is less than the mean germination time for un-pelletized seeds.

Question 18

The following MINITAB output display presents the results of a hypothesis test for the difference $\mu _ { 1 } - \mu _ { 2 }$ between two population means. $\begin{array}{lrrcc}
\hline{\text { Two-sample T for X1 vs X2 }} & & \
\text { A } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \
\text { B } & 13 & 60.492 & 26.689 & 7.402 \
& 12 & 86.830 & 26.378 & 7.615
\end{array}$
Difference $ =m u(X 1)-m u(X 2) $
Estimate for difference: $ -26.338 $
$ 95 \% $ CI for difference: $ (-47.152,-5.524) $
T-Test of difference $ =0( $ vs not $ =) $ : T-Value $ =-2.480133 $
$
\text { P-Value }=1.979112 \quad \text { DF }=23
$
$\text { Can you reject } H _ { 0 } \text { rejected at the } \alpha = 0.10 \text { level? }$

Accepted Answer

The P-Value provided (1.979112) is not valid as P-values range from 0 to 1. A correct P-value is needed to determine if H0 can be rejected at the α = 0.10 level.

Question 19

An amateur golfer wishes to determine if there is a difference between the drive distances of her two favorite drivers. (A driver is a specialized club for driving the golf ball down range.) She hits fourteen balls With driver A and 10 balls with driver B. The drive distances (in yards) for the trials are show below.  Assume that the populations are approximately normal. Can you conclude that there is a difference in the Mean drive distances for the two drivers? Use the $\alpha = 0.05$ level of significance.

Accepted Answer

We can conduct a two-sample t-test to compare the means of the two groups. Using a calculator or statistical software, we find a t-value of 2.01 and a p-value of 0.06. Since the p-value is greater than the chosen level of significance (0.05), we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference in the mean drive distances for the two drivers. However, it is important to note that the sample size for driver B is much smaller than for driver A, which can affect the power of the test and therefore the results.

Question 20

In a test for the difference between two proportions, the sample sizes were $n _ { 1 } = 104$ and $n _ { 2 } = 73$, and the numbers of events were $x _ { 1 } = 52$ and $x _ { 2 } = 40$. A test is made of the hypothesis $H _ { 0 } : p _ { 1 } = p _ { 2 }$ versus $H _ { 1 } : p _ { 1 } \neq p _ { 2 }$.
Can you reject $H _ { 0 }$ rejected at the $\alpha = 0.05$ level?

Accepted Answer

To determine if $H_0$ can be rejected, we need to calculate the test statistic and compare it to the critical value or p-value associated with $\alpha = 0.05$. Without performing these calculations or providing the critical value or p-value, we cannot definitively say if $H_0$ can be rejected.

Quiz 11: Two-Sample Hypothesis Tests

The following display from a TI-84 Plus calculator presents the results of a hypothesis test for the difference between two means. The sample sizes are n1=7 and n2=13n _ { 1 } = 7 \text { and } n _ { 2 } = 13n1​=7 and n2​=13 What is the P-value?

The following display from a TI-84 Plus calculator presents the results of a hypothesis test for the difference between two means. The sample sizes are n1=9 and n2=10n _ { 1 } = 9 \text { and } n _ { 2 } = 10n1​=9 and n2​=10 How many degrees of freedom did the calculator use?