Respuesta :
Answer:
(a) The value of z test statistics is -2.25.
(b) At 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.
(c) The p-value for this test statistic is 0.0122.
(d) We conclude that the average number of chocolate chips in all the cookies is less than 9.
Step-by-step explanation:
We are given that a sample of 16 cookies is taken to test the claim that each cookie contains at least 9 chocolate chips.
The average number of chocolate chips per cookie in the sample was 7.875. The standard deviation of the population is 2.
Let [tex]\mu[/tex] = the average number of chocolate chips in all the cookies.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 9 {means that the average number of chocolate chips in all the cookies is at least 9}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 9 {means that the average number of chocolate chips in all the cookies is less than 9}
The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean number of chocolate chips per cookie = 7.875
[tex]\sigma[/tex] = population standard deviation = 2
n = sample of cookies taken = 16
So, the test statistics = [tex]\frac{7.875-9}{\frac{2}{\sqrt{16} } }[/tex]
= -2.25
(a) The value of z test statistics is -2.25.
(b) Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.
Since our test statistic is less than the critical value of z as -2.25 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
(c) The P-value of the test statistics is given by;
P-value = P(Z < -2.25) = 1 - P(Z [tex]\leq[/tex] 2.25)
= 1 - 0.9878 = 0.0122
(d) Therefore, we conclude that the average number of chocolate chips in all the cookies is less than 9.
