Twenty-five adult males were involved in a study to determine the mean cholester
ID: 3150413 • Letter: T
Question
Twenty-five adult males were involved in a study to determine the mean cholesterol decrease when taking a new cholesterol drug. The drug is advertised to decrease total cholesterol on average by 40 points in 4 months. The observed mean cholesterol decrease in four months from the sample of 25 was 34 points and the standard deviation was 12.
What value must the test statistic be less than what value in order to reject the null hypothesis at the 1% significance level?
What is the value of the p-value in this case? Hint: The p-value is the tail area to the left of the test statistic associated with the test statistic. Based on the t-table a specific value of p cannot be stated, but rather p can be stated to be between two values.
What is the decision about the null hypothesis in this case with a significance level of 1%? Hint: If the p-value is less than the significance level then reject the null hypothesis.
What is the conclusion about the alternative hypothesis at the 1% significance level, based on the decision above about the null hypothesis?
What is the 99% confidence interval to estimate the parameter based on these data?
Does the above confidence interval contain the hypothesized parameter value (the value of 40 that is stated in the null hypothesis)?
Explanation / Answer
a)
What value must the test statistic be less than what value in order to reject the null hypothesis at the 1% significance level?
Thus, getting the critical t at 0.01 left tailed, by table/technology,
df = n - 1 = 24
tcrit = -2.492159473 [ANSWER]
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b)
What is the value of the p-value in this case? Hint: The p-value is the tail area to the left of the test statistic associated with the test statistic. Based on the t-table a specific value of p cannot be stated, but rather p can be stated to be between two values.
Formulating the null and alternative hypotheses,
Ho: u >= 40
Ha: u < 40
As we can see, this is a left tailed test.
Getting the test statistic, as
X = sample mean = 34
uo = hypothesized mean = 40
n = sample size = 25
s = standard deviation = 12
Thus, t = (X - uo) * sqrt(n) / s = -2.5
Also, the p value is, using a table.
0.005 < Pvalue < 0.01 [ANSWER]
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C)
What is the decision about the null hypothesis in this case with a significance level of 1%? Hint: If the p-value is less than the significance level then reject the null hypothesis.
As P < 0.01, we REJECT THE NULL HYPOTHESIS. [ANSWER]
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d)
What is the conclusion about the alternative hypothesis at the 1% significance level, based on the decision above about the null hypothesis?
Hence, we accept the null hypothesis. [ANSWER]
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e)
What is the 99% confidence interval to estimate the parameter based on these data?
As this is a left tailed test, we use an upper confidence bound.
Note that
Upper Bound = X + t(alpha) * s / sqrt(n)
where
alpha= (1 - confidence level)= 0.01
X = sample mean = 34
t(alpha) = critical t for the confidence interval = 2.492159473
s = sample standard deviation = 12
n = sample size = 25
df = n - 1 = 24
Thus,
Upper bound = 39.98118274
Thus, the confidence interval is
u < 39.98118274 [ANSWER, UPPER CONFIDENCE BOUND]
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f)
Does the above confidence interval contain the hypothesized parameter value (the value of 40 that is stated in the null hypothesis)?
As we should have expected, NO, 40 is not inside the interval. [ANSWER]
[It is not inside because we rejected the null hypothesis.]
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