Q:

A SAMPLE OF 16 FAMILIES WAS DRAWN FOR ANALYSIS OF THE AVERAGE GROCERY SPENDING. ASSUME THAT THE INDIVIDUAL WEEKLY RECORDS ARE NORMALLY DISTRIBUTED, WITH THE UNKNOWN POPULATION AVERAGE AND UNKNOWN POPULATION STANDARD DEVIATION. SAMPLE SUMMARIES WERE: (SAMPLE MEAN) = $154 AND (SAMPLE STANDARD DEVIATION) = $40. DO YOU HAVE SUFFICIENT EVIDENCE THAT THE POPULATION AVERAGE WOULD BE BELOW $172? USE THE SIGNIFICANCE LEVEL OF 10%.

Accepted Solution

A:
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.