a fast food outlet claims that the mean waiting time in line is less than 3.5 minutes. a random sample of 20…

Explanation:

Step1: State the hypotheses

$H_0:\mu\geq3.5$ (null hypothesis), $H_1:\mu < 3.5$ (alternative hypothesis)

Step2: Calculate the test - statistic

Since the population standard deviation is unknown, we use the t - test. The formula for the t - statistic is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$. Here, $\bar{x} = 3.7$, $\mu = 3.5$, $s = 0.8$, and $n = 20$. $t=\frac{3.7 - 3.5}{0.8/\sqrt{20}}\approx1.12$

Step3: Determine the degrees of freedom and critical value

The degrees of freedom $df=n - 1=20 - 1 = 19$. For a one - tailed test with $\alpha=0.05$ and $df = 19$, the critical value $t_{\alpha,df}=t_{0.05,19}=- 1.729$ (from t - distribution table).

Step4: Make a decision

Since the calculated $t = 1.12>-1.729$ (the critical value), we fail to reject the null hypothesis.

Answer:

We fail to reject the null hypothesis. There is not enough evidence to support the fast - food outlet's claim that the mean waiting time in line is less than 3.5 minutes at the $\alpha = 0.05$ significance level.