375 Cal
Statistical Methods
9.28
In this question we are trying to check the mean 18 ounces of weight is greater than the mean for our chosen sample of 18 whose mean is 17.78 and standard deviation 0.41. in this case we will use the Z table to test the null hypothesis. The first step is to state the null and the alternative hypothesis:
(a) Null hypothesis:
H0: X = 18
Alternative hypothesis:
Ha: X < 18
The other step is to determine our Z value from the table at the 5% test level, the value for 0.05 tests from the table is + 0r – 1.96
Our calculated Z value is as follows:
Z cal = X1 – X2/ (SD/ (n)½)
Where X1 is the sample mean, X2 is the stated mean, SD is the standard deviation and n is the sample size
Z cal = 17.78 – 18/ (0.41/(18)½)
Z cal = -2.2765
Z critical + or – 1.96
Because Z calculated lies in the rejection region we reject the null hypothesis that X = 18 and therefore we conclude that X < 18. Therefore the mean weight of the oats is less than 18 ounces.
(b) The level of test will determine whether we reject or accept a hypothesis, if the test was based on 1% level of test then the critical value would have been 2.57 and therefore we would have accepted our null hypothesis.
(c) The P valueis0.0113 while our alpha is 0.05 and because the p value is less than our alpha then this shows that the p value is consistent with our rejection of the null hypothesis.
9.29
In this question we are trying to check whether the time of waiting for a rental car which was estimated to have 19 minute mean in 2004 is less than our considered sample of 20 which showed a mean of 15 minutes and a standard deviation of 7 minutes
We state the null and the alternative hypothesis:
(a) Null hypothesis:
H0: X = 19
Alternative hypothesis:
Ha: X < 19
The other step is to determine our Z value from the table at the 5% test level, the value for 0.05 tests from the table is + 0r – 1.96
Our calculated Z value is as follows:
Z cal = X1 – X2/ (SD/ (n)½)
Where X1 is the sample mean, X2 is the stated mean, SD is the standard deviation and n is the sample size
Z cal = 15 – 19/ (7/(20)½)
Z cal = 2.55550626
Z critical + or – 1.96
Because Z calculated is greater than the critical Z we reject the null hypothesis that X = 19 and accept the alternative hypothesis that X < 19. Therefore the mean waiting time has decreased according to this hypothesis test.
(b) The P valueis0.0052 while our alpha is 0.05 and because the p value is less than our alpha then this shows that the p value is consistent with our rejection of the null hypothesis.
9.53
In this question we are trying to check whether Bob is a 90+ performer in his exams as his professor states, we will calculate the mean, the standard deviation of the given sample of his exams, his mean according to the professor is greater than 90 but according to the sample whose n = 8 the mean is 88.375 with a standard deviation of 4.9839, we test this hypothesis using the 1% test level.
(a) Null hypothesis:
H0: X = 90
Alternative hypothesis:
Ha: X < 90
The other step is to determine our Z value from the table at the 1% test level, the value for 0.01 tests from the table is + 0r – 2.576
Our calculated Z value is as follows:
Z cal = X1 – X2/ (SD/ (n)½)
Where X1 is the sample mean, X2 is the stated mean, SD is the standard deviation and n is the sample size
Z cal = 88.375 – 90/ (4.9839/(8)½)
Z cal = -0.922208326
Z critical + or – 2.576
Because Z calculated is less than Z critical we accept the null hypothesis that X = 90, therefore his performance is above 90 as the professor states
(b) The Z calculated value in this case lies in the acceptance zone and therefore we accept the null hypothesis that X = 90
(c) This test is based on some assumptions, these assumptions include that we assume that the sample is an unbiased random sample and the other assumption is that the sample is normally distributed.
(d) The P valueis0.1788 while our alpha is 0.01 and because the p value is greater than our alpha then this shows that the p value is consistent with our acceptance of the null hypothesis.
9.54
This question tests whether the mean of use of fax machines in the recent past in a firm has reduced for pages over 10 pages, a sample of 35 is collected and the mean is 14.44 with a standard deviation of 4.45 and the test is made at t1% level of test.
(a) Null hypothesis:
H0: X = 10
Alternative hypothesis:
Ha: X > 10
The other step is to determine our Z value from the table at the 5% test level, the value for 0.05 tests from the table is + 0r – 2.576
Our calculated Z value is as follows:
Z cal = X1 – X2/ (SD/ (n)½)
Where X1 is the sample mean, X2 is the stated mean, SD is the standard deviation and n is the sample size
Z cal = 10 – 14.44/ (4.45/(35)½)
Z cal = -5.902785222
Z critical + or – 2.576
Because Z calculated is greater than the critical value we reject the null hypothesis that X = 10 and therefore it is clear that the mean is greater than 10 pages.
(b). The P valueis less than 0.0001 while our alpha is 0.01 and because the p value is less than our alpha then this shows that the p value is consistent with our rejection of the null hypothesis.
References:
Lind and Mason (2004) Statistical Techniques, McGraw Hill,New York
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