in 1990 for all North Dakota residents was 13 minutes. A transportation resident
ID: 3313970 • Letter: I
Question
in 1990 for all North Dakota residents was 13 minutes. A transportation residents. Here ae data.ravel times, in minutes for a random sample of 35 North Dakota 51. The Census Bureau to work reau publication Census of population and Housing reports that the mean official obtained this y ear's t 29 400 12 10 6 41 25 21 5 4 19 27 10 8 3652 412 0 33 6217 218 38 2 13 8 14 11 2 At the 5% significant level, do the da ta provide sufficient evidence to conclude that the mean travel time to work all North Dakota residents has changed from the 1990 mean of 13 minu confidence interval method tes? Use 52. In an analysis investigation the usefulness of pennies, the cents portions of 80 randomly selected checks are recorded. The sample has a mean of 25.8 cents and standard 31.0 cents. If the amounts from 0 to 50 cents are all equally likely, the mean is expected to be 45.7 cents. Use a 0.02 significance level to test the claim that the sample is from a population with a mean less than 45.7 cents. Use p-value method deviation of 3. In an analysis investigation the usefulness of pennies, the cents portions of 80 randomly selected checks are recorded. The sample has a mean of 25.8 cents and standard devaon of 31.0 cents. If the amounts from 0 to 50 cents are all equally likely, the mean is expected to be 45.7 cents. Use a 0.02 significance level to test the claim that the sample is from a population with a mean less than 45.7 cents. Use p-value method.Explanation / Answer
Solution:-
53)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 45.7
Alternative hypothesis: < 45.7
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 0.02. 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).
SE = s / sqrt(n)
S.E = 3.466
DF = n - 1
D.F = 79
t = (x - ) / SE
t = - 5.74
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 - 5.74. We use the t Distribution Calculator to find P(t < - 5.74).
Thus the P-value in this analysis is less than 0.0001.
Interpret results. Since the P-value (almost 0) is less than the significance level (0.02), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that sample is from a population with a mean less than 45.7 cents.
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