Question

an online travel agent that specializes in cruises wanted to compare the cost of cruising to different destinations on various cruise lines. the following random sample data show the average cost of a seven - day cruise in a standard room to two locations. assume the population variances for the cruise fares for these destinations are equal. complete parts a through c below.
location 1 location 2
sample mean $991 $922
sample standard deviation $133 $125
sample size 9 13
a. choose the type of interval that should be found based on the information provided above.
oa. z
ob. non - pooled t
oc. pooled t
b. construct a 90% confidence interval to estimate the difference between the average rates of cruising to location 1 vs. cruising to location 2.
the 90% confidence interval is (□,□)
(round to two decimal places as needed.)

Explanation:

Part a

To determine the interval type, we analyze the given information:

• We have two independent samples (cruise costs to two locations).
• The population variances are assumed equal (stated in the problem).
• The sample sizes are small (\(n_1 = 9\), \(n_2 = 13\)) and population standard deviations are unknown (we have sample standard deviations).

For comparing two means with equal population variances and unknown population standard deviations, we use the pooled t - interval (pooled t - test/interval is appropriate when \(\sigma_1^2=\sigma_2^2\) and we estimate the common variance from the samples). A z - interval is used when population standard deviation is known (not the case here). A non - pooled t - interval is used when population variances are not assumed equal (contradicts the problem's assumption of equal variances). So the correct option is C.

Step 1: Calculate the pooled variance (\(s_p^2\))

The formula for pooled variance is \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)

Given: \(n_1 = 9\), \(s_1 = 133\), \(n_2 = 13\), \(s_2 = 125\)

First, calculate \((n_1 - 1)s_1^2=(9 - 1)\times133^2=8\times17689 = 141512\)

Then, calculate \((n_2 - 1)s_2^2=(13 - 1)\times125^2=12\times15625 = 187500\)

The denominator \(n_1 + n_2-2=9 + 13-2 = 20\)

So, \(s_p^2=\frac{141512+187500}{20}=\frac{329012}{20}=16450.6\)

Step 2: Calculate the standard error (\(SE\))

The formula for the standard error of the difference in means is \(SE = s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)

First, find \(s_p=\sqrt{16450.6}\approx128.26\)

Then, \(\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{9}+\frac{1}{13}=\frac{13 + 9}{117}=\frac{22}{117}\approx0.188\)

\(SE=128.26\times\sqrt{0.188}\approx128.26\times0.4336\approx55.62\)

Step 3: Calculate the difference in sample means (\(\bar{x}_1-\bar{x}_2\))

Given \(\bar{x}_1 = 991\), \(\bar{x}_2 = 922\)

\(\bar{x}_1-\bar{x}_2=991 - 922 = 69\)

Step 4: Find the t - critical value (\(t_{\alpha/2}\))

For a 90% confidence interval, \(\alpha=1 - 0.90 = 0.10\), \(\alpha/2=0.05\)

The degrees of freedom \(df=n_1 + n_2-2 = 20\)

From the t - distribution table, \(t_{0.05,20}\approx1.725\)

Step 5: Calculate the margin of error (\(E\))

The margin of error \(E=t_{\alpha/2}\times SE\)

\(E = 1.725\times55.62\approx95.94\)

Step 6: Calculate the confidence interval

The confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E\)

Substitute the values: \(69-95.94<\mu_1 - \mu_2<69 + 95.94\)

\(- 26.94<\mu_1 - \mu_2<164.94\) (rounded to two decimal places)

Answer:

C. Pooled t

Part b