Question

Two studies are conducted in York County.

a) A recent study stated that the average age of robbery victims was 63.5 years. A sample of 21 recent victims has a mean of 63.7 years and a standard deviation of 1.9 years. At the .05 level of significance, is the average age higher than originally believed? Use the 5 steps and complete a hypothesis test using the p-value method or critical value method.

b) A researcher wants to estimate the average weight loss obtained by patients at a residential weight loss clinic during the first week of a controlled diet and exercise regimen. How large a sample is needed to estimate this mean within 0.5 pounds with 95% confidence? Assume that these data are available from a preliminary observation of five individuals.

3.2       2.7       4.0       5.0       1.2

Answer 1

a)

Claim: The average age is higher than the original age. that is

The claim contains greater than sign so comes under alternative hypothesis and the null hypothesis contains less than or equal to or equal to sign.

Step 1: Null and alternative hypothesis:

Step 2: Test statistics

Here population Standard deviation is unknown so t-test is used.

The formula of t-test statistics

Where

s = standard deviation = 1.9

n - sample size = 21

Plug all the values in the formula of t-test statistics

Step 3: Critical value:

Alpha (level of significance) = 0.05

Degrees of freedom = n - 1 = 21- 1 = 20

Alternative hypothesis contains greater than sign so the test is right tailed test.

Using t distribution table the critical value for one-tailed area 0.05 with degrees of freedom 20 is 1.725

Step 4: Decision rule:

Decision rule: If Critical value > t-test statistics then fail to reject the null hypothesis otherwise reject the null hypothesis.

Here Critical value (1.725) is greater than test statistics (0.428) so we fail to reject the null hypothesis.

Step 5: Conclusion:

Here fail to reject the null hypothesis means in favor of null.

That is there is no sufficient evidence to support the claim that the average age is higher than 63.5 years.

b)

c = confidence level = 95% = 0.95

E = margin of error = 0.5

The formula to find the sample size,

where z - critical value at given confidence level and s - sample standard deviation

alpha = 1 - c = 1-0.95 = 0.05

alpha/2 = 0.025 => 1 - alpha/2 = 1 - 0.025 = 0.975

The z critical value uisng z table for area 0.975 is 1.96

The formula to find sample standard deviation is,

Where X - data values

Mean - average of all values and n - sample size which is 5

Subtract 3.22 from each value then take the square and add them.

Plug all the values in the formula of n,

Sample size = 32