(9) Consider the following hypotheses: H 0 : ? = 5,600 The population is normall
ID: 3430564 • Letter: #
Question
(9)
Consider the following hypotheses:
H0: ? = 5,600
The population is normally distributed with a population standard deviation of 620. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. Use Table 1.(Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)
Consider the following hypotheses:
Explanation / Answer
Answer to the question)
We got
standard deviaton s = 620
M = 5600
.
The formula of test statistic is :
z = (x bar - M ) / (s/sqrt(n))
.
Part a)
we got x bar = 5660
THus Ho: M = 5600
Ha: M > 5600
[right tailed test]
n = 145
.
On plugging the values we get
z = (5660-5600) / (620/sqrt(145))
z = 1.67
.
P value = 1 - 0.8790 = 0.1210 [this can be obtained from the z table]
Since the P value 0.1210 > alpha 0.10 , we FAIL to reject the null
.
Part b)
Again since the value of x bar is larger than M , we consider it to be a right tailed test
z = (5660 -5600) / (620/sqrt(325))
z = 1.74
P value =1- 0.9599 = 0.0401
P value 0.0401 < 0.10 , we REJECT the null hypothesis
.
part c)
In this case the vlaue of xbar 5310 < 5600 , hence we consider it to be a left tailed test
z = (5310 -5600) / (620/sqrt(33))
z = -2.69
Pvalue = 0.0036 .....[this value we can directly get from the z table , because the z table tells the left area]
Result: P value 0.0036 < alpha 0.10 , We REJECT the null
.
Part d)
it is again left tailed test
z = (5350-5600) / (620/sqrt(33))
z = -2.32
P value = 0.1017
Result : Since the P value 0.1017 > alpha 0.10 , We FAIL to reject the null
.
Part c)
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