modified . The battery of the was recently modified to last longer. A sam batter
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Question
modified . The battery of the was recently modified to last longer. A sam batteries had a mean life of 311 days with a standard deviation of 1 n is sus days. The lives of the at saple of 20 modification increase the mean life of the battery? (a) State the null hypothesis and the alternate hypothesis. (b) Show the decision rule graphically. Use the .05 significance level. (c) Compute the value of t. What is your decision regarding the null hypothesis marize your results. 9. Given the following hypotheses: Ho: 10 H1: > 10 A random sample of 10 observations is selected from a normal population. The sample mean was 12 and the sample standard deviation 3. Using the 05 significance level:. a. State the decision rule. b. Compute the value of the test statistic c. What is your decision regarding the null hypothesis? 10. Given the following hypotheses: H0: = 400 H1:+400 A random sample of 12 observations is selected from a normal population. The sample mean was 407 and the sample standard deviation 6. Using the.01 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? The Rocky Mountain district sales manager of Rath Publishing Inc., a coll lishing company, claims that the sales representatives make an average of 40 ege textbook pub- sales calls per 11. , thet thie estimate is too low To investiootExplanation / Answer
Solution:-
9)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 10
Alternative hypothesis: > 10
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 0.05. 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).
tcritical = 1.83
We will reject the H0, if t-value is greater than 1.83.
SE = s / sqrt(n)
S.E = 0.9487
DF = n - 1 = 10 - 1
D.F = 9
t = (x - ) / SE
t = 2.11
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 2.11.
Thus the P-value in this analysis is 0.03
Interpret results. Since the P-value (0.03) is less than the significance level (0.05), we have to reject the null hypothesis.
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