value: 1.25 points At the time she was hired as a server at the Grumney Family R
ID: 2925217 • Letter: V
Question
value: 1.25 points At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, "You can average $83 a day in tips." Assume the population of daily tips is normally distributed with a standard deviation of $4.07. Over the first 45 days she was employed at the restaurant, the mean daily amount of her tips was $84.86. At the 0.05 significance level, can Ms. Brigden conclude that her daily tips average more than $83? a. State the null hypothesis and the alternate hypothesis. 0H0' -83 : H-83 b. State the decision rule. OReject H, if z > 1.65 Reject Ho if z1.65 OReject H, if z1.65 Reject Ho if z > 1.65 c. Compute the value of the test statistic. (Round your answer to 2 decimal places.) Value of the test statistic d. What is your decision regarding Ho? Do not reject Ho O Reject Ho e. What is the p-value? (Round your answer to 4 decimal places.) p-value References eBook & ResourcesExplanation / Answer
given that n=45 , x=$84.86 and s= 4.07
The null hypothesis is that tips are less than or equal to $83, "H0: Mean <= 83". The alternative is that they are more: "HA: Mean > 83".
answer is option 3
test statistic, "(observed mean - hypothesized mean) / (standard dev. / ?n)", and compare it to a critical value of the standard normal distribution. For ?=0.05, this critical value is 1.645. If |z| > 1.65, we will reject the null hypothesis.
answer is option 4
The test statistic is:
z =(84.86 - 83) / (4.07 / ?45)
= 3.07.
Since the test statistic is greater than the critical value (3.07 > 2.33),
answer is we REJECT the null hypothesis, in favor of the alternative (that the mean is greater than $83.00).
The p-value is: P[z > 3.07] =1- P[z <3.07]
= 1-0.99893
=0.00
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