Question

# A clinic offers a weight-reduction program. A review of its records revealed the following weight losses,...

A clinic offers a weight-reduction program. A review of its records revealed the following weight losses, in pounds, for a
random sample of 29 of its clients after the program:
6 14 0 5 3 2 3 0 3 0
5 7 0 4 8 7 6 5 7 6
8 0 0 0 4 2 0 5 3
Goodness-of-Fit Test
Shapiro-Wilk W Test

W Prob&lt;W
0.899470 0.0095*
Note: Ho = The data is from the Normal distribution. Small p-values reject Ho.
a) Test the claim that the data is from a normal population by using the results from the Shapiro Wilk test. Use a 5% level of
significance.
Ho: The data IS from a normal population
Ha: The data is NOT from a normal population
(1) Decision Rule: If the p-value (Prob &lt; W) is more than α = 0.05, then we would reject Ho / not reject Ho (select one)
Test statistic: (1) the p-value = prob &lt; W=
Decision: (1) Reject Ho Do not reject Ho (Select one)
Conclusion in plain English: (1) Select one of the following
i) We have strong evidence that the data is not from a normal distribution
ii) We have strong evidence that the data is from a normal population
iii) We have no evidence that the data is not normally distributed
iv) none of the above
b) (4) Calculate the mean and standard deviation of this sample.

c) (11) Test the claim that the true population standard deviation is 5. Use a 2% level of significance.
Ho: Ha:
Decision Rule: Test statistic:

Decision Conclusion

d) (1) Based on your conclusion to (c) do you need to set up a confidence interval for the population variance? Yes/No
e) (3) Irrespective of your conclusion in (c), set up a 98% confidence interval for the population variance.

f) (3) Using your conclusion to (c), set up a 98% confidence interval for the population mean.

(a)

(1) Test statistics(W)=0.89947 and p-value(of test statistics) = 0.009515

Here, at 5% level of significance, p-value(0.009515), which is less than 0.05 and in this case, we reject the null hypothesis(Ho).

Decision: Reject Ho

Conclusion: We have strong evidence that the data is not from a normal distribution.

(b) mean of given sample= 3.896552 and standard deviation of sample= 3.352449

(c)

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