Question

A prescription medicine is supposed to contain an average of 250 parts per million (ppm) of a certain chemical. If the concentration is higher or lower than this, the drug may not be effective. The manufacturer runs a check to see if the mean concentration of the chemical in a large shipment conforms to the target level of 250 ppm or not. If it does not, the manufacturer will stop production (costing time and money) to remedy the situation. A simple random sample of 100 portions of the shipment is selected, and the mean concentration in these portions is found to be 247 ppm with a standard deviation of 12 ppm.

A. Conduct a hypothesis test to determine if the mean concentration of the drug in all portions of this shipment conforms to the target level of 250 ppm or not (Since 99 degrees of freedom is NOT on the t-hypothesis table, use the p-value associated with the largest number of degrees of freedom on the table, which is 60)

B. In the context of this problem, identify a type 1 error.

C. Discuss the consequences of a type 1 error.

D. In the context of this problem, identify a type 2 error.

F. Discuss the consequences of a type 2 error.

G. Estimate, with 90% confidence, the mean concentration of the drug in all portions of the shipment (From the extended t-table, the critical t-value for 90% confidence is 1.66)

H. Determine whether or not the procedures in parts A and G are valid

Answer 1

SOLUTION A: (1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 250

Ha: μ ≠ 250

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the critical value for a two-tailed test is tc​=1.66.

The rejection region for this two-tailed test is R={t:∣t∣>1.66}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=2.5>tc​=1.66, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0141, and since p=0.0141<0.1, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 250, at the 0.1 significance level.

SOLUTION B: TYPE I ERROR IS REJECTING H0 WHEN IT IS TRUE. REJECTING THAT prescription medicine contains an average of 250 parts per million (ppm) of a certain chemical and concluding that  prescription medicine does not contains an average of 250 parts per million (ppm) of a certain chemical.

SOLUTION C: CONSEQUENCES OF TYPE 1 ERROR IS the manufacturer will stop production (costing time and money) to remedy the situation.

SOLUTION D: TYPE 2 ERROR IS ACCEPTING H0 WHEN IT IS FALSE.  ACCEPTING THAT prescription medicine contains an average of 250 parts per million (ppm) of a certain chemical and concluding that  prescription medicine does contains an average of 250 parts per million (ppm) of a certain chemical.

SOLUTION E: CONSEQUENCE OF TYPE 2 ERROR IS MANUFACTURER KEEP ON MANUFACTURING SUCH PRODUCT AND WASTING TIME AND MONEY.

SOLUTION F: sample Mean = 247
t critical = 1.66
sM = √(122/100) = 1.2

μ = M ± t(sM)
μ = 247 ± 1.66*1.2
μ = 247 ± 1.99

90% CI [245.01, 248.99].

You can be 90% confident that the population mean (μ) falls between 245.01 and 248.99.

SOLUTION G: PROCEDURES IN PART A AND G ARE VALID BECAUSE BOTH ARE CONCLUDING THE SAME RESULT.