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Answer :
a) The 95% confidence interval for the mean amount of the active chemical in the drug is (97.06 mg, 102.54 mg).
b) At a significance level of α = 0.05, the null hypothesis that the population mean amount of the active chemical in the drug is 95 mg is rejected.
a) To calculate the 95% confidence interval for the mean amount of the active chemical in the drug, we can use the formula:
Confidence Interval = sample mean ± (critical value) * (standard deviation / √sample size)
Since we want a 95% confidence interval, the critical value corresponds to a 2.5% level of significance on each tail of the distribution. For a sample size of 31, the critical value can be obtained from a t-table or calculator. Assuming a normal distribution, the critical value is approximately 2.039.
Confidence Interval = 99.8 mg ± (2.039) * (7 mg / √31)
Confidence Interval = (97.06 mg, 102.54 mg)
Therefore, we can be 95% confident that the true mean amount of the active chemical in the drug lies within the interval of (97.06 mg, 102.54 mg).
b) To test the null hypothesis that the population mean amount of the active chemical in the drug is 95 mg, we can use a t-test. With a sample mean of 99.8 mg and a known standard deviation of 7 mg, we can calculate the t-value:
t = (sample mean - hypothesized mean) / (standard deviation / √sample size)
t = (99.8 mg - 95 mg) / (7 mg / √31)
t ≈ 2.988
At a significance level of α = 0.05, and with 30 degrees of freedom (sample size minus 1), the critical t-value can be found from a t-table or calculator. The critical t-value is approximately 1.699.
Since the obtained t-value (2.988) is greater than the critical t-value (1.699), we reject the null hypothesis. This means that there is evidence to suggest that the population mean amount of the active chemical in the drug is different from 95 mg at a significance level of 0.05.
Learn more about Null hypothesis here:
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