he charitable foundation for a large metropolitan hospital is conducting ealthy
ID: 3316718 • Letter: H
Question
he charitable foundation for a large metropolitan hospital is conducting ealthy individuals; the average annual donor income in the most recent survey was andom sample of 200 current donors showed a mean annual income of $103,157 and a standard deviation of $27,498. a study to characterize its donor base. In the past, most donations have come from relatively right at $100,000. The foundation believes the average has now increased. A (A) Specify a hypothesis test to test the foundation's claim (B) Compared to the most recent survey, test your hypothesis at the 10% level (C) Report the p-value for this test by showing your sample mean, standard e or degrees of freedom and test statistic. (D) Report on whether we accept or reject the nul hypothesis, and if it is rejected is it a convincing rejection, a moderately strong rejection, or a so-so For the toolbar, press ALT+F10 (PC) or ALT+FN F10 (Mac) T T T Arial ' 3(12px) T·-·-· A) Ho M S100,000 and HA M>100,000 .ees C) p value 053 D)Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 100,000
Alternative hypothesis: > 100,000
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 0.10. 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 = 1944.4
DF = n - 1
D.F = 199
t = (x - ) / SE
t = 1.62
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.
The observed sample mean produced a t statistic test statistic of 1.62.
Thus the P-value in this analysis is 0.0526
Interpret results. Since the P-value (0.0526) is smaller than the significance level (0.10), we have to reject the null hypothesis.
From the above test we have strong evidence in the favor of the claim that averages has increased.
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