5. Give the null hypothesis and the alternative hypothesis used here.
Null hypothesis: Proportion of washing machines that are still working is 90% Alternate hypothesis: Proportion of washing machines that are still working is not 90%
6. Give the P-value and the conclusion to the question from this P-value.
Z-statistic = (0.85 – 0.90)/SQRT(0.9*0.1/100) = -1.6667 P-value = 0.09558
As the p-value is greater than the 0.05 level of significance, we fail to reject the null hypothesis. Thus, the data doesn’t contradict the maker’s claim at 0.05 level of significance.
For problems 7 and 8, an auto manufacturer claims that the average length of time that one of its cars is owned before it requires a major repair is at least seven years. Assume that a survey of ten owners of the manufacture's cars finds that they went an average of 6 years before a major repair and the sample standard deviation for such time lengths was 1.8 years. Use the data to test the manufacture's claim at a 5% significance level www.statisticsassignmenthelp.com
7. Give the null hypothesis and alternative hypothesis and critical value(s) used to test this claim.
Null Hypothesis: average length of time that one of the cars is owned before it requires a major repair is greater than or equal to seven years
Alternate Hypothesis: average length of time that one of the cars is owned before it requires a major repair is less than seven years
8. Give the test statistic and use this to test the manufacturer's claim at a 5% significance level and draw a conclusion based on this.
Test statistic = (6 – 7)(1.8/sqrt(10)) = -1.7568 P-value = 0.0564 > 0.05
Therefore, at 5% significance level, we fail to reject the null hypothesis.There is insufficient evidence to reject the manufacturer’s claim
For problems 9 and 10, a machine shop claims that the measurements of the parts it produces are uniform with a standard deviation of 1.3 millimeters. A test sample of 31 parts from the shop has a standard deviation of 1.8 millimeters.
9. Give the null and alternative hypothesis and critical value(s). www.statisticsassignmenthelp.com
Null hypothesis: The variance of parts is equal to 1.32 = 1.69 millimeters2
Alternate hypothesis:The variance of parts is not equal to 1.3 = 1.69 millimeters2
The critical values are 18.493 and 43.733
10. Give the test statistic and use this to find if this result contradicts the shop's claim at a 10% level of significance.
Test statistic = (N-1)*s2/σ2= (30)*(1.82)/(1.32) = 57.514
As, the test statistic is beyond the critical values, we reject the null hypothesis. We conclude the standard deviation of measurement of parts is not 1.3 millimeters. www.statisticsassignmenthelp.com