It is reported that 20% of a certain type of holiday light set will fail if the
ID: 3249102 • Letter: I
Question
It is reported that 20% of a certain type of holiday light set will fail if the temperature dips to 0 degree F. In a test, each of a random sample of 100 such light sets was subjected to a temperature of 0 degree F. Let X be the count of the tested lights sets that fail. Let p be the actual proportion of holiday lights sets that will fail when subjected to a temperature of 0 degree F. We believe that that actual proportion of those light sets that will fail when subjected to a temperature of 0 degree F is greater than 20%, so we will make this our test (alternative) hypothesis. Upon performing the temperature test on 100 holiday lights sets we discover the 27 of them fail. (a) State the specific hypotheses that will be tested. (b) What is the point estimate for p? (c) What is the test statistic for this hypothesis test? (type and value) (d) What is the p-value for this hypothesis test?Explanation / Answer
Solution:-
14)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 95
Alternative hypothesis: 95
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. 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 = 0.40
DF = n - 1 = 16 - 1
D.F = 15
t = (x - ) / SE
t = - 2.5
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.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 15 degrees of freedom is less than - 2.5 or greater than 2.5.
Thus, the P-value = 0.025
Interpret results. Since the P-value (0.0245) is greater than the significance level (0.01), we cannot reject the null hypothesis.
From the above test we have sufficient evidecne in the favor of the claim that = 95.
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