Part a) Find the test static and p-value. Decide if your reject or fail to rejec
ID: 3374627 • Letter: P
Question
Part a) Find the test static and p-value. Decide if your reject or fail to reject null hypothesis. A health administration recommends that individuals consume 900 mg of calcium daily. After an advertising campaign aimed at male teenagers, a dairy association states that male teenagers consume more than the recommended daily amount of calcium. To support this statement, the association obtained a random sample of 40 male teenagers and found that the mean amount of calcium consumed was 925 mg, with a standard deviation of 100 mg. Is there significant evidence to support the statement of the association at the ?-0.05 level of significance? Choose the correct hypotheses. ??: 900 H1: ? > 900 Find the test statistic. (Round to two decimal places as needed.) Part b) Find the test static and p-value. Decide if your reject or fail to reject null hypothesis. The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is108.8 85.3 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table. Complet parts (a) and (b) below 69.9 56.5 75.0 66.5 82. 94.4 85.1 71.5 80.2 (b) Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of ?=0.1 First determine the appropriate hypotheses. Ho: ? 85.3 1: H 85.3 Find the test statistic. (Round to two decimal places as needed.)Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u < 900
Alternative hypothesis: u > 900
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 15.8114
DF = n - 1
D.F = 39
t = (x - u) / SE
t = 1.58
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 1.58.
Thus the P-value in this analysis is 0.061.
Interpret results. Since the P-value (0.061) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that there is significant evidence to support the statemnet of the association.
Do not reject the null hypothesis.