Answer to Question #231427 in Statistics and Probability for Kanwal

Question

To compare customer satisfaction levels of two competing ice cream companies, 8 customers of Company 1 and 5 customers of Company 2 were randomly selected and were asked to rate their ice creams on a five-point scale, with 1 being least satisfied and 5 most satisfied. The survey results are summarized in the following table. Construct an interval estimate at 95% confidence interval for difference of means of two ice cream companies b. Write the number of the correct option.

1. The 95% CI is (0.6685 to 1.3385)

2. The 95% CI is (0.6685 to 1.3115)

3. The 95% CI is (0.6415 to 1.3385)

4. The 95% CI is (0.6685 to 2.3385)

5. The 95% CI is (1.2146 to 1.3115)

Company 1

N1=8

Mean value of sample 1=3.21

standard deviation of sample 1=0.31

Company 2

N2=5

Mean value of sample 2=2.22

standard deviation of sample 2=0.21

Answer 1

Solution:

Given:

n₁ = 8

$\bar x_1 = 3.21$

s₁ = 0.31

n₂ = 5

$\bar x_2 = 2.22$

s₂ = 0.21

Assuming both variances are equal:

Since samples are less than 30 in this problem we are dealing with t-distirbution with n₁ + n₂ – 2 = 8 + 5 – 2 = 11 degrees of freedom.

The table t-value for a 95% confidence interval with 25 df is t_{0.025, 11} = 2.201

The formula for a 95% confidence interval for the difference of two population means:

$$(\bar x_1-\bar x_2)\pm t_{0.025, 11}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

where

$$s_p=\sqrt{\frac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}}$$

$s_p=\sqrt{\frac{0.31^2\cdot7+0.21^2\cdot4}{8+5-2}}=0.28$

Confidence interval:

$(3.21-2.22)\pm2.201\cdot0.28\sqrt{\frac{1}{8}+\frac{1}{5}}=0.99\pm0.35$

We are 95% confident that the difference in the two population means is between 0.64 and 1.34. Zero is not in this interval so there is a significant difference of means of two ice cream companies.

Constructing a confidence interval for the difference of means when both variances are not equal:

$$(\bar x_1-\bar x_2)\pm t\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$

where df is the smaller of n₁–1 and n₂–1

df = min(8-1, 5-1)=4

Therefore the table t-value for a 95% confidence interval with 4 df is t_{0.025, 4} = 2.776

$(3.21-2.22)\pm 2.776\sqrt{\frac{0.31^2}{8}+\frac{0.21^2}{5}}=0.99\pm 0.4$

We are 95% confident that the difference in the two population means is between 0.59 and 1.39. Zero is not in this interval so there is a significant difference of means of two ice cream companies.