Question 8 Trinity 2010A charity believes that 42 of all donors f.pdf
Question 8 The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use alpha = 0.10. Complete parts (a) through (d). Using a chi-square goodness-of- fit test, you can decide, with some degree of certainty, whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as follows. H0: The test scores have a normal distribution. Ha: The test scores do not have a normal distribution. Find the expected frequencies. (Round to the nearest integer as needed.) Determine the critical value, ,and the rejection region. Choose the correct rejection region below. Calculate the test statistic. chi2 = (Round to three decimal places as needed.) Decide whether to reject or fail to reject the null hypothesis. Then interpret the decision in the context of the original claim. H0. At the 10% significance level, there enough evidence to reject the claim that the test scores are normally distributed. Solution (a) observedexpectedO - E(O - E).
![Question 8 Trinity 2010
A charity believes that 42% of all donors from the previous year will donate again in the current
year. A sample of 300 donors was taken.
a) What is the standard deviation of the sample proportion that will donate again?
b) What is the probability that more than half of these sample members will donate again?
c) Construct a 99% acceptance interval for the proportion of sample members that will donate
again
Suppose that the sample proportion of last years donors who donated again this year is . Do you
think the charity is correct to believe that 42% of all donors from last year will donate again?
Why (not)?
Solution
Given n=300, p=0.42
(a) The standard deviation=[p*(1-p)/n] = sqrt(0.42*(1-0.42)/300) =0.02849561
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(b) The probability is
P(phat>0.5)
= P((phat-p)/[p*(1-p)/n] > (0.5-0.42)/sqrt(0.42*(1-0.42)/300))
=P(Z> 2.81)
= 0.0025 (check standard normal table)
--------------------------------------------------------------------------------------------------------------
(c) a=0.01, |Z(0.005)|=2.58 (check standard normal table)
So 99% CI is
p ± Z*[p*(1-p)/n]
--> 0.42 ±2.58*sqrt(0.42*(1-0.42)/300))
--> ( 0.3464813, 0.4935187)](https://image.slidesharecdn.com/question8trinity2010acharitybelievesthat42ofalldonorsf-230327035752-775c49b6/85/Question-8-Trinity-2010A-charity-believes-that-42-of-all-donors-f-pdf-1-320.jpg)
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Question 8 Trinity 2010A charity believes that 42 of all donors f.pdf
- 1. Question 8 Trinity 2010 A charity believes that 42% of all donors from the previous year will donate again in the current year. A sample of 300 donors was taken. a) What is the standard deviation of the sample proportion that will donate again? b) What is the probability that more than half of these sample members will donate again? c) Construct a 99% acceptance interval for the proportion of sample members that will donate again Suppose that the sample proportion of last years donors who donated again this year is . Do you think the charity is correct to believe that 42% of all donors from last year will donate again? Why (not)? Solution Given n=300, p=0.42 (a) The standard deviation=[p*(1-p)/n] = sqrt(0.42*(1-0.42)/300) =0.02849561 -------------------------------------------------------------------------------------------------------------- (b) The probability is P(phat>0.5) = P((phat-p)/[p*(1-p)/n] > (0.5-0.42)/sqrt(0.42*(1-0.42)/300)) =P(Z> 2.81) = 0.0025 (check standard normal table) -------------------------------------------------------------------------------------------------------------- (c) a=0.01, |Z(0.005)|=2.58 (check standard normal table) So 99% CI is p ± Z*[p*(1-p)/n] --> 0.42 ±2.58*sqrt(0.42*(1-0.42)/300)) --> ( 0.3464813, 0.4935187)