A factory that manufactures bolts is performing a quality control experiment. Each object should have a length of no more than 1
1 answer:
The test statistic z is 2.33, the p-value corresponding to the test statistic z value is 0.0099.
<h3>What is Probability?</h3>Probability is the measure of the likeliness of happening of an event.
The mean of the data is 15.07 centimeters
The standard deviation of the data is 0.26 centimeters.
n = 75
Significance Level ∝ = 0.01
According to the null and alternative hypothesis
Hₐ : ≤15 vs H₁ :
>15
The test statistic z is given by
Z = ( 15.07 -15)/(0.26/√75)
Z = 2.33
The p-value corresponding to z value is 0.0099
as p-value < significance level, therefore the H₁ : >15 is acceptable.
No, significant evidence is not present to tell that the length of bolts is actually greater than the mean value at a significance level of 0.01.
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79 is a prime, so we cannot reduce it. 
Answer:
We conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.
Step-by-step explanation:
We are given that in a certain school district, it was observed that 32% of the students in the element schools were classified as only children (no siblings).
However, in the special program for talented and gifted children, 135 out of 347 students are only children.
<u><em>Let p = population proportion of only children in the special program.</em></u>
So, Null Hypothesis, : p = 32% {means that the proportion of only children in the special program is equal to the proportion for the school district}
Alternate Hypothesis, : p
32% {means that the proportion of only children in the special program is significantly different from the proportion for the school district}
The test statistics that would be used here <u>One-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of only children in the special program =
= 0.39
n = sample of students = 347
So, <em><u>test statistics</u></em> =
= 2.673
The value of z test statistics is 2.673.
<em>Now, at 0.02 significance level the z table gives critical value of -2.3263 and 2.3263 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.