The local oil changing business is very busy on Saturday mornings and is considering expanding.A national study of similar businesses reported the mean number of customers waiting to have their oil changed on Saturday morning is 3.6.Suppose the local oil changing business owner, wants to perform a hypothesis test.The null hypothesis is the population mean is 3.6 and the alternative hypothesis that the population mean is not equal to 3.6.The owner takes a random sample of 16 Saturday mornings during the past year and determines the sample mean is 4.2 and the sample standard deviation is 1.4.It can be assumed that the population is normally distributed.The observed "t" value for this problem is _______.
A)0.05
B)0.43
C)1.71
D)1.33
E)0.71

Question

Quizplus · Accepted Answer

The observed "t" value is calculated using the formula: $t = \frac{\bar{x} - \mu}{(s/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values: $t = \frac{4.2 - 3.6}{(1.4/\sqrt{16})} = \frac{0.6}{(1.4/4)} = \frac{0.6}{0.35} = 1.71$.

The Local Oil Changing Business Is Very Busy on Saturday