Suppose the number of free throws in a basketball game by one player are normally distributed with a standard deviation 0.97 fre
1 answer:
Answer: 1.960
Step-by-step explanation:
The value of z we use to calculate a confidence interval with a () confidence level is a two-tailed test value i.e. represented by :-
Given : The level of confidence:
Then, significance level :
With the help of standard normal distribution table for z , we have
Hence, the value of z should be used to calculate a confidence interval with a 95% confidence level =1.960
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Answer:
The null hypothesis is .
The alternate hypothesis is
The pvalue of the test is 0.0182 < 0.05, which means that there is sufficient evidence to show that the manufacturer needs to recalibrate the machines.
Step-by-step explanation:
A manufacturer must test that his bolts are 4.00 cm long when they come off the assembly line.
This means that the null hypothesis is that they are 4.00 cm long, that is:
At the alternate hypothesis we test if it is different. So
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
4 is tested at the null hypothesis:
This means that
After sampling 196 randomly selected bolts off the assembly line, he calculates the sample mean to be 4.14 cm.
This means that
He knows that the population standard deviation is 0.83 cm.
This means that
Value of the test statistic:
Pvalue of the test and decision:
The pvalue of the test is the probability that the sample mean is different by at least 4.14 - 4 = 0.14 from the target value, which is P(|z| > 2.36), which is 2 multiplied by the pvalue of z = -2.36.
Looking at the z-table, z = -2.36 has a pvalue of 0.0091
2*0.0091 = 0.0182
The pvalue of the test is 0.0182 < 0.05, which means that there is sufficient evidence to show that the manufacturer needs to recalibrate the machines.