Null and alternate hypothesis
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1.A machine produces 5-inch nails. A sample of 12 nails was selected and their lengths determined. The results are as follows:
4.94 4.94 5.07 4.91 4.98 5.00 4.94 5.09 5.00 4.92 5.08 5.06
Assuming that a = 0.10, test the hypothesis that the population mean is equal to 5.
?State the null and alternate hypotheses
?Calculate the mean and standard deviation
?Determine which test statistic applies, and calculate it
?Determine the critical value(s).
?State your decision: Should the null hypothesis be rejected?
7. The pulse rates below were recorded over a 30-second time period, both before and after a physical fitness regimen. The data is shown below for 8 randomly selected participants. Is there sufficient evidence to conclude that a significant amount of improvement took place? Assume pulse rates are normally distributed. Test using a = 0.05.
?State the null and alternate hypotheses
?Calculate the mean and standard deviation
?Determine which test statistic applies, and calculate it
?Determine the critical value(s).
?State your decision: Should the null hypothesis be rejected?
Before 43 34 37 41 41 59 36 58
After 37 36 40 49 52 51 31 55
(Points :10)
8. You are given the following data. Test the claim that there is a difference in the means of the two groups. Use a = 0.05.
Group A Group B
xbar1 = 10 xbar2 = 4
s1 = 15 s2 = 19
n1 = 89 n2 = 69
?State the null and alternate hypotheses
?Determine which test statistic applies, and calculate it
?Determine the critical value(s).
?State your decision: Should the null hypothesis be rejected?
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Solution Summary
A null and alternative hypothesis is examined.
Solution Preview
Please see the attached file.
1. The null hypothesis to be tested is
Against the alternative hypothesis
Given a sample of 12 observations
Mean = = 4.99
Variance = = 0.0044
Standard deviation = = 0.0663
The test statistic , this statistic follows student's t distribution with n-1 degrees of freedom.
The calculated value of the test statistic
From student's t tables with 11 d.f the critical value for = 0.10 is =1.80
(Note that this critical value can be obtained by using TINV ...
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