Question

The weights of the apples grown on an orchard are normally distributed. The mean weight has been m0 = 9.500 ounces. Because of the inclement weather this year, the farmer would like to know if the mean weight has decreased. A random sample of n = 16 apples is selected. The sample mean is x = 9.320 ounces and the sample standard deviation is s = 0.18 ounces. A test needs to be conducted to test if the population mean has decreased at a significance level a = 0.05 .
1. The null and alternative hypotheses should be_____
2. Using the action limits to set up the decision rule, the decision rule should be______
3. The test statistic of this test is ______
4. The conclusion of this test should be _________

Answer 1

Answer:

The answers to the questions are;

1. null hypothesis is μ = 9.5

   alternative hypothesis is μ < 9.5

2. The null hypothesis is μ = 8.96

   alternative hypothesis is μ < 8.96

3. The test statistic for the test t = -4

4. We accept the null hypothesis that the mean remain the same and we reject the alternative hypothesis that the mean has changed.

Step-by-step explanation:

To solve the question, we note that the farmer would like to know if the apples have growing smaller at the current season therefore;

1. The null hypothesis is μ = 9.5

The alternative hypothesis is μ < 9.5

2. Action limits are the minimum and maximum values of acceptability and it s derived by

Action limit = Average ± 3 × Standard deviation

= 9.5 ± 3 × 0.18 = Upper limit = 10.04

                              Lower limit = 8.96

Using the action limit for decision rule since we have inclement weather, our limit should be the lower limit of 8.96 and the null and alternative hypothesis becomes

The null hypothesis is μ = 8.96

The alternative hypothesis is μ < 8.96

3. The test statistic is given by

$t = \frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }$

Where:

x' = Sample mean = 9.320

μ = Population mean = 9.500

σ = Standard deviation = 0.18

n = Number of data = 16

Therefore t = $\frac{9.32-9.500}{\frac{0.18}{\sqrt{16} } }$ = -4

The test statistic for the test t = -4

4. The conclusion of the test is

At a significance level a = 0.05, z$_{0.05}$ = 1.645 from normal distribution tables

Therefore since the test statistic is -4 we accept it where z < -z$_{0.05}$

hence since -4 < -1.645, we accept the null hypothesis that the mean remain the same.