Answer:
Yes, we have sufficient evidence that the population average would be below $172.
Step-by-step explanation:
Let X be the random variable that represents an individual weekly record of grocery spending. We have observed n = 16 grocery spendings and we have the sample summaries [tex]\bar{x}[/tex] = $154 and s = $40. We know that X is normally distributed.
We have the following null and alternative hypothesis
[tex]H_{0}: \mu = 172[/tex] vs [tex]H_{1}: \mu < 172[/tex] (lower-tail alternative)
We will use the test statistic
[tex]T = \frac{\bar{X}-172}{S/\sqrt{16}}[/tex] and the observed value is
[tex]t = \frac{154-172}{40/\sqrt{16}} = -1.8[/tex]
if [tex]H_{0}[/tex] is true, then T has a t distribution with n-1 = 15 degrees of freedom.
We will use the significance level of 10%, the rejection region is determined by the 10th quantile of the t distribution with 15 degrees of freedom. This value is -1.3406 and the rejection region is {t | t < -1.3406}. Because the observed value t = -1.8 is less than -1.3406, i.e., because the observed value belongs to the rejection region, we reject the null hypothesis at the level of significance of 10%. Therefore, we have sufficient evidence that the population average would be below $172.
