Respuesta :
Answer:
The p-value of her test is 0.15386.
Step-by-step explanation:
We are given that a buyer for a grocery chain inspects large truckloads of apples to determine the proportion p of apples in the shipment that are rotten.
She selects a simple random sample of 200 apples from the over 20000 apples on the truck and the sample contains 9 rotten apples.
Let p = proportion of of apples in the shipment that are rotten.
SO, Null Hypothesis, [tex]H_0[/tex] : p = 0.06 {means that the proportion of apples in the shipment that are rotten is equal to 0.06}
Alternate Hypothesis, [tex]H_A[/tex] : p < 0.06 {means that the proportion of apples in the shipment that are rotten is less than 0.06}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = proportion of apples that are rotten in a sample of 200 apples = [tex]\frac{9}{200}[/tex] = 0.045
n = sample of apples = 200
So, test statistics = [tex]\frac{0.045 -0.06}{{\sqrt{\frac{0.045(1-0.045)}{200} } } } }[/tex]
= -1.02
The value of the test statistics is -1.02.
Now, P-value of the test statistics is given by;
P-value = P(Z < -1.02) = 1 - P(Z [tex]\leq[/tex] 1.02)
= 1 - 0.84614 = 0.15386
Hence, the p-value of her test is 0.15386.
Using the z-distribution, it is found that the p-value for her test is of 0.1867.
The null hypothesis is:
[tex]H_0: p = 0.06[/tex]
The alternative hypothesis is:
[tex]H_a: p < 0.06[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
For this problem, the parameters are:
[tex]p = 0.06, n = 200, \overline{p} = \frac{9}{200} = 0.045[/tex]
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.045 - 0.06}{\sqrt{\frac{0.06(0.94)}{200}}}[/tex]
[tex]z = -0.89[/tex]
The p-value is found using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, with z = -0.89, and is of 0.1867.
You can learn more about the use of the z-distribution for an hypothesis test at https://brainly.com/question/25912188
