Question 1

$\text { Find the required sample size without making a guess for the observed value of } \hat { p } \text {. }$
-A manufacturer of kitchen utensils wishes to estimate the proportion of left handed people in the population. Obtain a sample size that will ensure a margin of error of at most 0.046 for a 93%
Confidence interval.

Accepted Answer

To find the sample size without an initial guess for $\hat{p}$, use $\hat{p} = 0.5$ for maximum variability. The formula for sample size $n$ is $n = \left(\frac{Z \cdot \sqrt{0.5 \cdot 0.5}}{E}\right)^2$, where $Z$ is the Z-score for 93% confidence (approximately 1.81) and $E$ is the margin of error (0.046). Calculating this gives approximately 388.

Question 2

Use the one-proportion z-test to perform the specified hypothesis test. Use the critical-value approach.
-$\mathrm { x } = 30 , \mathrm { n } = 169 , \mathrm { H } _ { 0 } : \mathrm { p } = 0.11 , \mathrm { H } _ { \mathrm { a } } : \mathrm { p } \neq 0.11 , \alpha = 0.05$

Accepted Answer

The correct z-score calculation for the given data ($x = 30, n = 169, p = 0.11$) results in a z-score of 2.81. Given the two-tailed test with $\alpha = 0.05$, the critical z-values are $\pm 1.96$. Since $2.81 > 1.96$, we reject $H_0$.

Question 3

Suppose that you wish to estimate a population proportion and want to determine a sample size that will ensure a given margin of error for a $95 \%$ confidence interval. Suppose further that an educated guess of $0.3$ is used for $\hat { p }$ when determining the sample size. What values of the observed value of $\hat { p }$ will yield a larger margin of error than the one specified?

Accepted Answer

The margin of error is maximized when $\hat{p}(1-\hat{p})$ is maximized, which occurs at $\hat{p} = 0.5$. Thus, values of $\hat{p}$ between 0.3 and 0.7 will yield a larger margin of error than the one specified with $\hat{p} = 0.3$.

Question 4

$\text { Find the required sample size without making a guess for the observed value of } \hat { p } \text {. }$
-A researcher wishes to estimate the proportion of all drivers who exceed the speed limit on a certain stretch of road where accidents frequently happen. Obtain a sample size that will ensure a
Margin of error of at most 0.021 for a 98% confidence interval.

Accepted Answer

The sample size needed to achieve a margin of error of 0.021 for a 98% confidence interval is 3078. This can be calculated using the formula: n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level, p is the estimated proportion of drivers who exceed the speed limit, q is the complement of p (1-p), and e is the margin of error. Since we do not have an estimate for p, we can use the conservative estimate of p = 0.5, which will give us the largest possible sample size. Plugging in the values, we get: n = (2.326^2 * 0.5 * 0.5) / 0.021^2 = 3078.

Question 5

Find the indicated margin of error.
-In a survey of 9500 T.V. viewers, 2850 said they watch network news programs. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

Accepted Answer

The margin of error (ME) for a proportion is calculated using the formula: ME = Z * sqrt[(p(1-p))/n], where Z is the Z-score for the confidence level (2.576 for 99%), p is the sample proportion (2850/9500), and n is the sample size (9500). Plugging in these values gives the margin of error closest to 0.0121.

Question 6

Use the two-proportion plus-four z-interval procedure to find the required confidence interval.
-A survey of randomly chosen adults found that 27 of the 86 women and 37 of the 87 men follow regular exercise programs. Construct a 95% confidence interval for the difference in the
Proportions of women and men who have regular exercise programs.

Accepted Answer

The two-proportion plus-four z-interval procedure involves adjusting the sample sizes and successes for both groups by adding two successes and two failures to each. This adjustment is used to calculate the confidence interval for the difference in proportions when sample sizes are small or when the observed proportions are close to 0 or 1. The formula for the confidence interval is:$$ \left( \hat{p}_1 - \hat{p}_2 \right) \pm z^* \sqrt{ \frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2} } $$where $\hat{p}_1$ and $\hat{p}_2$ are the adjusted proportions of successes in each group, $n_1$ and $n_2$ are the adjusted sample sizes, and $z^*$ is the critical value from the standard normal distribution for the desired level of confidence (approximately 1.96 for 95% confidence).Given:- Original counts for women: 27 successes out of 86 (exercise), adjusted to 29 out of 90.- Original counts for men: 37 successes out of 87 (exercise), adjusted to 39 out of 91.Adjusted proportions:- Women: $\hat{p}_1 = \frac{29}{90}$- Men: $\hat{p}_2 = \frac{39}{91}$The calculation involves substituting these values and the critical value into the formula. The correct interval, based on these calculations, is the one that includes the difference in proportions between women and men, accounting for the variability in those proportions. The correct answer reflects the interval that would result from these calculations, which is option C, indicating a confidence interval for the difference in proportions that includes both negative and positive values, suggesting that the true difference could favor either group or be negligible.

Question 7

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval.
-n = 138, x = 86; 90% level

Accepted Answer

The answer of The number of successes and the sample...

Question 8

Find the P-value for the indicated hypothesis test.
-A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 39% of them plan to go into
General practice. Find the P-value for a test of the school's claim.

Accepted Answer

The null hypothesis is that the proportion is 28%, and the alternative hypothesis is that it is more than 28%. Using a one-sample z-test for proportions, the test statistic is calculated, and the P-value is found to be 0.0026, which corresponds to the probability of observing a sample proportion of 39% or more if the true proportion is 28%.

Question 9

A two-proportions z-test is to be performed using the P-value approach. Use the given sample data to find the P-valuefor the hypothesis test.
-$\mathrm { x } _ { 1 } = 41 , \mathrm { n } _ { 1 } = 100 , \mathrm { x } _ { 2 } = 35 , \mathrm { n } _ { 2 } = 140 ; \mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = \mathrm { p } _ { 2 } , \mathrm { H } _ { \mathrm { a } } : \mathrm { p } _ { 1 } > \mathrm { p } _ { 2 } , \alpha = 0.10$

Accepted Answer

The correct P-value for the given two-proportions z-test, based on the provided sample data and the hypothesis test, is $0.0043$. This is calculated using the formula for the test statistic in a two-proportions z-test and then finding the corresponding P-value for the one-tailed test since the alternative hypothesis is directional ($p_1 > p_2$).

Question 10

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-interval procedure is appropriate.
-x = 77, n = 120

Accepted Answer

The one-proportion z-interval procedure is appropriate if the sample size is large enough and the success-failure condition is met. Both np and n(1-p) should be greater than 10. Here, p = x/n = 77/120. So, np = 77 and n(1-p) = 43, both of which are greater than 10, making the procedure appropriate.

Question 11

The number of successes and the sample size are given for a simple random sample from a population. Determine thesample proportion, $\hat{ p. }$
-x = 16, n = 25

Accepted Answer

The sample proportion $\hat{p}$ is calculated as the number of successes (x) divided by the sample size (n), so $\hat{p} = \frac{x}{n} = \frac{16}{25} = 0.64$.

Question 12

Determine whether the following statement regarding a hypothesis test for two population proportions. However small the difference between two population proportions, for sufficiently large sample sizes, the null hypothesis of equal population proportions is likely to be rejected.

Accepted Answer

As sample sizes increase, the standard error of the difference between two proportions decreases, making even small differences between the proportions statistically significant, leading to the rejection of the null hypothesis of equal proportions.

Question 13

The numbers of successes and the sample sizes are given for independent simple random samples from two populations.Use the two-proportions z-interval procedure to obtain the specified confidence interval.
-$\mathrm { x } _ { 1 } = 32 , \mathrm { n } _ { 1 } = 80 , \mathrm { x } _ { 2 } = 17 , \mathrm { n } _ { 2 } = 20.99 \%$ confidence interval

Accepted Answer

The answer of The numbers of successes and the sample...

Question 14

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-test is appropriate.
-$x = 60 , n = 120 , H _ { 0 } : p = 0.6 , H _ { a } : p < 0.6$

Accepted Answer

The one-proportion z-test is appropriate here because the sample size is large enough (n = 120) to assume the distribution of the sample proportion is approximately normal (np = 72 and n(1-p) = 48, both greater than 10), which meets the requirements for conducting a one-proportion z-test.

Question 15

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion plus-four z-interval procedure to find the required confidence interval.
-n = 188, x = 157; 95% level

Accepted Answer

The answer of The number of successes and the sample...

Question 16

Suppose that you wish to estimate a population proportion and want to determine a sample size that will ensure a given margin of error for a $95 \%$ confidence interval. Suppose further that an educated guess is not available for $\hat { \mathrm { p } }$ and that $0.25$ is used for the value of $\hat { \mathrm { p } } ( 1 - \hat { \mathrm { p } } )$ when determining the sample size. True or false, if the observed value of $\hat { \mathrm { p } }$ is not equal to $0.5$, then the sample size will be larger than necessary?

Accepted Answer

The answer of Suppose that you wish to estimate a...

Question 17

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-interval procedure is appropriate.
-x = 86, n = 90

Accepted Answer

The answer of The number of successes and the sample...

Question 18

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval.
-$$\mathrm { n } = 85 , \mathrm { x } = 41 ; 98 \% \text { level }$$

Accepted Answer

The sample proportion $\hat{p} = \frac{x}{n} = \frac{41}{85}$. For a 98% confidence interval, the z-value is approximately 2.33. Using the formula for the confidence interval $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, the interval is calculated to be approximately 0.356 to 0.608.

Question 19

Assume that you wish to estimate a population proportion, p. For the given margin of error and confidence level,determine the sample size required.
-You wish to estimate the proportion of shoppers that use credit cards. Obtain a sample size that will ensure a margin of error of at most 0.01 for a 95% confidence interval. It is deemed reasonable
To presume that of those sampled, the percentage using credit cards will be at least 56%.

Accepted Answer

To determine the sample size required for estimating a population proportion with a specific margin of error and confidence level, we use the formula: $n = \frac{Z^2 \times p(1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the margin of error. For a 95% confidence interval, the Z-score is approximately 1.96. Given that $p = 0.56$ and $E = 0.01$, we substitute these values into the formula: $n = \frac{(1.96)^2 \times 0.56(1-0.56)}{(0.01)^2}$, which calculates to approximately 9466, rounding to the nearest whole number.

Question 20

A hypothesis test is to be performed for a population proportion. For the given sample data and null hypothesis, compute the value of the test statistic, $z = \frac { \hat { p } - \mathrm { p } _ { 0 } } { \sqrt {  { p } _ { 0 } \left( 1 - \mathrm { p } _ { 0 } \right) / \mathrm { n } } }$
-A research group wants to determine whether the proportion of car accidents caused by drivers using cell phones has changed from the previous value of 13%. They obtained 10,000 auto accident
Reports and found that 14% were caused by drivers using cell phones. The hypotheses are $\mathrm { H } _ { 0 } : \mathrm { p } = 0.13 , \mathrm { H } _ { \mathrm { a } } : \mathrm { p } \neq 0.13$ 
where p is the proportion of car accidents caused by drivers using cell
Phones.

Accepted Answer

The test statistic $z$ is calculated using the formula given, with $\hat { p } = 0.14$, $p_0 = 0.13$, and $n = 10000$. Substituting these values into the formula gives $z = \frac { 0.14 - 0.13 } { \sqrt { 0.13 \times (1 - 0.13) / 10000 } } = \frac { 0.01 } { \sqrt { 0.13 \times 0.87 / 10000 } } = \frac { 0.01 } { \sqrt { 0.1131 / 10000 } } = \frac { 0.01 } { \sqrt { 0.00001131 } } = \frac { 0.01 } { 0.003364 } \approx 2.974$.

Quiz 12: Inferences for Population Proportions

Use the one-proportion z-test to perform the specified hypothesis test. Use the critical-value approach. - $\mathrm { x } = 30 , \mathrm { n } = 169 , \mathrm { H } _ { 0 } : \mathrm { p } = 0.11 , \mathrm { H } _ { \mathrm { a } } : \mathrm { p } \neq 0.11 , \alpha = 0.05$

Find the indicated margin of error. -In a survey of 9500 T.V. viewers, 2850 said they watch network news programs. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval. -n = 138, x = 86; 90% level

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-interval procedure is appropriate. -x = 77, n = 120

The number of successes and the sample size are given for a simple random sample from a population. Determine thesample proportion, $\hat{ p. }$ -x = 16, n = 25

Determine whether the following statement regarding a hypothesis test for two population proportions. However small the difference between two population proportions, for sufficiently large sample sizes, the null hypothesis of equal population proportions is likely to be rejected.

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-test is appropriate. - $x = 60 , n = 120 , H _ { 0 } : p = 0.6 , H _ { a } : p < 0.6$

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion plus-four z-interval procedure to find the required confidence interval. -n = 188, x = 157; 95% level

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-interval procedure is appropriate. -x = 86, n = 90

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval. - $\mathrm { n } = 85 , \mathrm { x } = 41 ; 98 \% \text { level }$

Quiz 12: Inferences for Population Proportions

Find the indicated margin of error. -In a survey of 9500 T.V. viewers, 2850 said they watch network news programs. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval. -n = 138, x = 86; 90% level

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-interval procedure is appropriate. -x = 77, n = 120

The number of successes and the sample size are given for a simple random sample from a population. Determine thesample proportion, p.^\hat{ p. }p.^​ -x = 16, n = 25

Determine whether the following statement regarding a hypothesis test for two population proportions. However small the difference between two population proportions, for sufficiently large sample sizes, the null hypothesis of equal population proportions is likely to be rejected.

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-test is appropriate. - x=60,n=120,H0:p=0.6,Ha:p<0.6x = 60 , n = 120 , H _ { 0 } : p = 0.6 , H _ { a } : p < 0.6x=60,n=120,H0​:p=0.6,Ha​:p<0.6

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion plus-four z-interval procedure to find the required confidence interval. -n = 188, x = 157; 95% level

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-interval procedure is appropriate. -x = 86, n = 90

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval. - n=85,x=41;98% level \mathrm { n } = 85 , \mathrm { x } = 41 ; 98 \% \text { level }n=85,x=41;98% level

The number of successes and the sample size are given for a simple random sample from a population. Determine thesample proportion, $\hat{ p. }$ -x = 16, n = 25

The number of successes and the sample size are given for a simple random sample from a population. Decide whetherusing the one-proportion z-test is appropriate. - $x = 60 , n = 120 , H _ { 0 } : p = 0.6 , H _ { a } : p < 0.6$

The number of successes and the sample size are given for a simple random sample from a population. Use theone-proportion z-interval procedure to find the required confidence interval. - $\mathrm { n } = 85 , \mathrm { x } = 41 ; 98 \% \text { level }$