Question

1. WHO is examining the improvement in life expectancy in an underdeveloped country. Historical data for the population shows a population mean life expectancy of 58.3 years, and with a population standard deviation of 4.52 years. A recent sample of 75 people produced a sample mean of 60.8 years and a sample standard deviation of 2.84 years. Evaluate H₀: μ ≤ 58.3 and H_A: μ > 58.3 at a .01 level of significance.

2. A medical research team at the Johns Hopkins University is investigating the effectiveness of a new medical treatment for a rare disease. Data indicates that under the current treatment, patients have a mean white blood cell count of 22 cells/cc (cubic centimeter). This distribution is understood to be normally distributed, but the population standard deviation is not known. The new treatment will only be deemed effective if it significantly increases the patient’s white blood cell count above 22 cells/cc.
   

The team has applied the treatment to a random sample of 10 patients. They have reported the following blood cell counts: 21, 25, 18, 24, 19, 25, 22, 20, 27, 24.

At α = .01, is the new treatment effective?

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 58.3
Alternative hypothesis: u > 58.3

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

S.E = 0.521925
z = (x - u) / SE

z = 4.79

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesised population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 4.79.

P-value = P(z > 4.79)

Use the z-value calculator for finding p-values.

P-value = 0.00

Interpret results. Since the P-value (0.00) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is improvement in life expectancy in an underdeveloped country.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 22
Alternative hypothesis: u > 22

Note that these hypotheses constitute a one-tailed test.

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).

S.E = 0.93393
DF = n - 1

D.F = 9
t = (x - u) / SE

t = 0.5354

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesised population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 0.5354.

P-value = P(t > 0.5354)

Use the t-value calculator for finding p-values.

P-value = 0.303

Interpret results. Since the P-value (0.303) is greater than the significance level (0.01), we failed to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the new treatment is effective.

SE n

Mean Sum S.D Count S.E Blood cells count 22.5 225 2.9533408577782 10 0.9339283817414

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