Question

In 1998, the average price for bananas was 51 cents per pound. In 2003, the following 7 sample prices (in cents) were obtained from local markets:

50, 53, 55, 43, 50, 47, 58.

Is there significant evidence to suggest that the average retail price of bananas is different than 51 cents per pound? Test at the 5% significance level.

Answer 1

`t=(50.857-51)/((5.014)/(√(7)))=-0.075`

The degrees of freedom are given by:

`df=n-1=7-1=6`

The p value for this case would be given:

`p_v = 2*P(t_6 <-0.075)=0.943`

The p value for this case is lower than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 51

Explanation:

Info given

50, 53, 55, 43, 50, 47, 58.

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

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

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

represent the mean height for the sample

`s=5.014` represent the sample standard deviation for the sample

`n=7` sample size

represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value

Hypothesis to test

We want to test if the true mean is equal to 51, the system of hypothesis would be:

Null hypothesis:
`\mu = 51`

Alternative hypothesis:
`\mu \\eq 51`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing we got:

`t=(50.857-51)/((5.014)/(√(7)))=-0.075`

The degrees of freedom are given by:

`df=n-1=7-1=6`

The p value for this case would be given:

`p_v = 2*P(t_6 <-0.075)=0.943`

The p value for this case is lower than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 51