A pharmaceutical company has developed a new drug to reduce cholesterol. A regulatory agency will recommend the new drug for use if there is convincing evidence that the mean reduction in cholesterol level after one month of use is more than 20 milligrams/deciliter (mg//dl), because a mean reduction of this magnitude would be greater than the mean reduction for the current most widely used drug. The pharmaceutical company collected data by giving the new drug to a random sample of 50 people from the population of people with high cholesterol. The reduction in cholesterol level after one month of use was recorded for each individual in the sample, resulting in a sample mean of bar(x)=24mg//dl and a standard deviation of s_(x)=15mg//dl. Perform a significance test at a 5% significance level.

Question

A pharmaceutical company has developed a new drug to reduce cholesterol. A regulatory agency will recommend the new drug for use if there is convincing evidence that the mean reduction in cholesterol level after one month of use is more than 20 milligrams/deciliter  (mg//dl), because a mean reduction of this magnitude would be greater than the mean reduction for the current most widely used drug. The pharmaceutical company collected data by giving the new drug to a random sample of 50 people from the population of people with high cholesterol. The reduction in cholesterol level after one month of use was recorded for each individual in the sample, resulting in a sample mean of  bar(x)=24mg//dl and a standard deviation of  s_(x)=15mg//dl. Perform a significance test at a  5% significance level.

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State Hypotheses: State the null hypothesis (H0) and the alternative hypothesis (H1). H0: μ = 20 mg/dl (The mean reduction in cholesterol level is equal to 20 mg/dl) H1: μ > 20 mg/dl (The mean reduction in cholesterol level is greater than 20 mg/dl)Set Significance Level: Set the significance level (α). The significance level is given as 5%, so α = 0.05.Calculate Test Statistic: Calculate the test statistic using the sample mean, population mean under the null hypothesis, sample standard deviation, and sample size. The test statistic (t) is calculated using the formula: t = (bar(x) - μ) / (s_x / sqrt(n)) where bar(x) is the sample mean, μ is the population mean under H0, s_x is the sample standard deviation, and n is the sample size. t = (24 - 20) / (15 / sqrt(50)) t = 4 / (15 / 7.071) t = 4 / 2.121 t ≈ 1.886Determine Degrees of Freedom: Determine the degrees of freedom (df) for the t-distribution. The degrees of freedom is equal to the sample size minus one. df = n - 1 df = 50 - 1 df = 49Find Critical Value: Find the critical value from the t-distribution table for a one-tailed test at the 5% significance level with 49 degrees of freedom. The critical value (t_critical) for df = 49 at α = 0.05 for a one-tailed test is approximately 1.677 (this value may vary slightly depending on the t-distribution table used).Compare Test Statistic: Compare the test statistic to the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis. Since t ≈ 1.886 and t_critical ≈ 1.677, and t > t_critical, we reject H0.Draw Conclusion: Draw a conclusion based on the comparison. Since we have rejected the null hypothesis, there is convincing evidence at the 5% significance level to suggest that the mean reduction in cholesterol level after one month of use of the new drug is more than 20 mg/dl.

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