AaBbCcDdEe AaBbCcDdEe AaBbCcDc AaBbCcDdEE Normal No Spacing Heading 2 Styles Pan

Question

AaBbCcDdEe AaBbCcDdEe AaBbCcDc AaBbCcDdEE Normal No Spacing Heading 2 Styles Pane Heading 1 Title Subtitle 2. To see if young men ages 8-17 years spend a different amount than the national average of $24.44 per shopping trip to a local mall, the manager surveyed 30 young men. She found that the average amount spent per trip was $23.37 with a standard deviation of $3.70. With alpha 0.05, can it be concluded that 8-17 years old spend a different amount at the local mall than the national average? a) State Ho b) State H c) Is this a one-tailed test or a two-tailed test? Why? d) What are the rejection regions for this specific test? Why? (It may help to draw it) e) In a nutshell, what is the researcher trying to provide evidence for? f) Compute the relevant test statistic (consult slides for formula) g) Compute the p value from the previous step h) Based on this value, do we reject or fail to reject Ho i) State your findings using this template: "Based on the findings, there is/is not enough evidence to suggest that (state Ha), p Te Focus 100% -

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

a) Null hypothesis: = 24.44
b) Alternative hypothesis: 24.44

c) Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it 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 = 0.6755

DF = n - 1 = 30 - 1

D.F = 29
t = (x - ) / SE

t = 1.58

tcritical = + 2.045

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 29 degrees of freedom is less than -1.58 or greater than 1.58.

Thus, the P-value = 0.124.

Interpret results. Since the P-value (0.124) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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