Question

The weights of the watermelons grown on a farm are normally distributed. The mean weight has been μ0=32.5 pounds. Because of the use of a new fertilizer, the farmer would like to know if the mean weight has changed. A random sample of n=9 watermelons is selected. The sample mean is x=33.8 pounds and the sample standard deviation is s=2.45 pounds. Test if the population mean has changed at a significance level a=0.05.

Answer 1

`t=(33.8-32.5)/((2.5)/(√(9)))=1.56`

`p_v =2*P(z>1.56)=0.157`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean of weights for thr watermellons is not different from 32.5 5% of signficance.

Explanation:

Data given and notation

`\bar X=33.8` represent the sample mean

`s=2.45` represent the sample standard deviation

`n=9` sample size

`\mu_o =32.5` represent the value that we want to test

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

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 32.5, the system of hypothesis would be:

Null hypothesis:
`\mu =32.5`

Alternative hypothesis:
`\mu \\eq 32.5`

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(33.8-32.5)/((2.5)/(√(9)))=1.56`

P-value

The degrees of freedom are given by
`df = n-1= 9-1=8`

Since is a two-sided test the p value would be:

`p_v =2*P(z>1.56)=0.157`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean of weights for thr watermellons is not different from 32.5 5% of signficance.