Question

The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 383​ 285​ 300​ 300 A consumer organization reports that the mean price of 7.5-cubic-foot refrigerators is greater than $300. Do the data provide convincing evidence of this claim? Use the α = 0.05 level of significance and assume the population is normally distributed.

Answer 1

`t=(310.9-300)/((31.09)/(√(10)))=1.109`

The degrees of freedom are given by:

`df=n-1=10-1=9`

And the p value would be given by:

`p_v =P(t_(9)>1.109)=0.148`

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

`\bar X =(\sum_(i=1)^n X_i)/(n)`

`s =\sqrt{(\sum_(i=1)^n (x_i -\bar X)^2)/(n-1)}`

`\bar X=310.9` represent the sample mean

`s=31.09` represent the standard deviation for the sample

`n=10` sample size

`\mu_o =300` represent the value to test

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 300`

Alternative hypothesis:
`\mu > 300`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(310.9-300)/((31.09)/(√(10)))=1.109`

The degrees of freedom are given by:

`df=n-1=10-1=9`

And the p value would be given by:

`p_v =P(t_(9)>1.109)=0.148`

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.