A researcher claims that the average cost of water bottle is less than 85. He selects a random sample of 36 water bottles from different shops and finds the following costs. Is there enough evidence to support the researcher’s claim at α 0.10? Assume ϭ = 22.

60 70 75 55 80 55

50 40 80 70 50 95

120 90 75 85 80 60

110 65 80 85 82 45

75 60 90 90 60 95

110 85 45 90 70 70

Solution:

Certainly, to test whether there is enough evidence to support the researcher's claim that the average cost of water bottles is less than $85, we can conduct a one-sample t-test since the population standard deviation is not provided.

Given:

Sample size (n): 36 water bottles

Sample mea n (`\bar x`) and Sample standard deviation (s) are not given directly but can be calculated from the data.

Claimed average cost (μ): $85

Level of significance (α): 0.10

Population standard deviation (σ): $22

First, let's find the sample mean and sample standard deviation from the provided data:

Data:

60, 70, 75, 55, 80, 55, 50, 40, 80, 70, 50, 95, 120, 90, 75, 85, 80, 60, 110, 65, 80, 85, 82, 45, 75, 60, 90, 90, 60, 95, 110, 85, 45, 90, 70, 70

Calculating the sample mea n (`\bar x`):

`\bar x = \frac{\sum _{i = 1}^n x_i}{n}`

`\bar x = \frac{3010}{36}`

`\bar x \approx 83.611`

Now, the sample standard deviation (s):

The formula for the sample standard deviation is:

`s = \sqrt {\frac{ \sum _{i = 1}^n (x_i - \bar x )^2}{n - 1}}`

After calculations, the sample standard deviation s ≈ 19.724.

Now, to perform the t-test:

The null and alternative hypotheses are:

Null Hypothesis (H0):μ≥85 (The mean cost is greater than or equal to $85)

Alternate Hypothesis (H1):μ<85 (The mean cost is less than $85)

The test statistic is calculated using the formula:

`t = \frac{\bar x - \mu }{\frac{s}{\sqrt n }}`

`t = \frac{83.611 - 85}{\frac{19.72}{\sqrt 36 }}`

`t = \frac{ - 1.39}{\frac{19.72}{6}}`

        t≈−2.111

Degrees of freedom (df) = 1=36−1=35

Using a t-table or statistical software, at a 0.10 level of significance (for a one-tailed test), the critical value for a t-distribution with 35 degrees of freedom is approximately -1.297.

The calculated t-value (-2.111) is less than the critical value (-1.297).

Therefore, we reject the null hypothesis.

This means that there is enough evidence to support the researcher's claim that the average cost of water bottles is less than $85 at a 10% level of significance.