Question

17 multiple choice 1 point
conduct the appropriate hypothesis test and compute the test statistic.

the industry standard for the delivery time on a pizza is 30 minutes. the manager of a pizza shop periodically tests to make sure they are performing better than the industry standard. the most recent random sample of 11 delivery times yielded a mean delivery time of 27 minutes with a sample standard deviation of 6.

assuming delivery times are normally distributed, perform the appropriate hypothesis test at a 0.05 significance level in order to determine whether there is sufficient sample evidence to conclude that their average delivery time is less than 30 minutes.

no, because the test statistic is -1.66
yes, because the test statistic is -0.5
cannot be determined
no, because the test statistic is -0.5
yes, because the test statistic is -1.66

Explanation:

Step1: Identify Test Type

This is a one - sample t - test (since population standard deviation is unknown, we use sample standard deviation \(s\)). The null hypothesis \(H_0:\mu = 30\) and the alternative hypothesis \(H_1:\mu<30\). The formula for the t - test statistic is \(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Plug in Values

We know that \(\bar{x} = 27\), \(\mu=30\), \(s = 6\), and \(n = 11\).
First, calculate the denominator \(\frac{s}{\sqrt{n}}=\frac{6}{\sqrt{11}}\approx\frac{6}{3.3166}\approx1.809\).
Then, calculate the numerator \(\bar{x}-\mu=27 - 30=- 3\).
Now, calculate the t - statistic: \(t=\frac{-3}{1.809}\approx - 1.66\).

Step3: Determine Significance

The significance level \(\alpha = 0.05\), and for a one - tailed test with \(n - 1=10\) degrees of freedom, the critical value \(t_{\alpha,df = 10}\) (from t - table) is approximately \(- 1.812\) (since it's a left - tailed test). Our calculated test statistic \(t=-1.66\) is greater than the critical value \(- 1.812\) (i.e., \(-1.66>-1.812\)), so we fail to reject the null hypothesis? Wait, no, wait. Wait, the alternative hypothesis is \(\mu < 30\), so we reject \(H_0\) if \(t < t_{\alpha}\). Wait, \(t_{\alpha}\) for left - tailed test with \(\alpha = 0.05\) and \(df = 10\) is \(- 1.812\). Our \(t=-1.66\) is greater than \(-1.812\), so we do not reject \(H_0\)? But wait, the options: Let's re - check the calculation. Wait, maybe I made a mistake in critical value. Wait, the sample size is \(n = 11\), so \(df=n - 1 = 10\). The critical value for a one - tailed test at \(\alpha=0.05\) is \(t_{0.05,10}=- 1.812\) (from t - distribution table). Our test statistic is \(t=-1.66\). Since \(-1.66>-1.812\), we fail to reject \(H_0\), which means there is not sufficient evidence to conclude that the average delivery time is less than 30 minutes. But wait, the last option says "Yes, because the test statistic is - 1.66". Wait, no, maybe I messed up the critical value. Wait, no, let's recalculate the test statistic. \(\bar{x}=27\), \(\mu = 30\), \(s = 6\), \(n = 11\). \(t=\frac{27 - 30}{\frac{6}{\sqrt{11}}}=\frac{-3}{\frac{6}{3.3166}}=\frac{-3}{1.809}\approx - 1.66\). Now, for a left - tailed test, the p - value: the p - value is \(P(T < - 1.66)\) with \(df = 10\). From t - table, \(P(T < - 1.66)\) is between \(0.05\) and \(0.10\) (since \(t_{0.10,10}=-1.372\) and \(t_{0.05,10}=-1.812\)). Since \(\alpha = 0.05\), and p - value>\(\alpha\) (because \(-1.66\) is greater than \(-1.812\), so the area to the left of \(-1.66\) is greater than \(0.05\)), we fail to reject \(H_0\). But the option "No, because the test statistic is - 1.66" is an option. Wait, but let's check the options again. Wait, maybe I made a mistake in the critical value direction. Wait, the alternative hypothesis is \(\mu<30\), so it's a left - tailed test. The critical value is \(t_{\alpha}=-1.812\). Our test statistic is \(t=-1.66\), which is in the non - rejection region (since \(-1.66>-1.812\)), so we do not reject \(H_0\), meaning there is not sufficient evidence. So the correct option is "No, because the test statistic is - 1.66".

Answer:

No, because the test statistic is - 1.66