..ooo AT&T; 11:00 PM @(? 47%. + HW10 HW 10 (1 point) The nicotine content in cig
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Question
..ooo AT&T; 11:00 PM @(? 47%. + HW10 HW 10 (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) and standard deviation 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a random sample of 15 cigarettes of this brand. The sample yields an average of 1.6 mg of nicotine. Conduct a test using a significance level of -0.05 (a) The test statistic (b) The critical value, 20 (c) The final conclusion is A. There is sufficient evidence to support the claim. B. There is not sufficient evidence to support the claim Courses Calendar To Do Notifications MessagesExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 1.5
Alternative hypothesis: > 1.5
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
SE = 0.02582
z0 = 1.645
z = (x - ) / SE
z = 3.87
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 z statistic test statistic of 3.87. We use the z Distribution Calculator to find P(z > 3.87).
Thus the P-value in this analysis is 0.00005
Interpret results. Since the P-value (0.00005) is less than the significance level (0.05), we have to reject the null hypothesis.
There is sufficient evidence to support the claim.
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