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Sigma Known – population mean hypothesis test

Practical application scenario

Sigma Known – population mean hypothesis test

Sampling distribution’s Standard deviation can be calculated directly where standard deviation is know and the normal distribution that is standardized can be used to obtain a z multiple. This is easily done by using NORMSINV.  The p-value for every 3 potential test conditions can be calculated, and then compared with every alpha level to observe whether to reject the null hypothesis.

The inputs include Population mean of the Hypothesis, sigma (the standard deviation of the population),  n as the sample size , at X-bar (sample mean) 295 &It – where the right number of the situation is put , 50It – where the right number of situation is put, and 298 &It – where the right number of situation is put.

Estimation standard error calculations: 1.697

Statistic of the test (z) is 1.532

The results shows that for the given level of Alpha, H0 equals Alpa:  (0.01) one tailed, no rejection, H0: Mu is > 294, p is 0.94 one tailed,

H0: Mu &It; 294, p is 0.063 two tailed, no rejection, H0: Mu is 294, p is 0.126 no rejection.

 The estimate of the situation and at the given alpha level, H0 must be 0.05

 

Sigma unknown – population mean hypothesis test

The standard deviation of the sample can be used as an estimate I calculating sampling distribution’s standard deviation as the estimate’s standard error.  T distribution is used to obtain a multiple that corresponds the required confidence level. This is by use of TINV Excel function. The calculation of p for every 3 likely test conditions can be done and then compared to every alpha level to decide whether to reject the null hypothesis.

This involves; 1.1 &It – put the right situation number: 61&It – put the right situation number; 7.3 &It put the right situation number;

Standard error is 0.1356

t(test statistics) is 1.8443

d.f (degree of freedom: 59)

 

  • no rejection

0.5         no rejection

One tailed

H0:Mu   &gt  = 7 

P-value = 0.965

One tailed

H0: Mu < =7,      no rejection

p -value = 0.035    rejected

One tailed

H0: Mu = 7   > no rejection

p- Value = 0.07         > no rejection

The situation must be 0.1 not rejected

 

 

Scenario 2

The sampling distribution of standard deviation is calculated using sample proportion hypothesis tests and normal distribution that is standardized  is used to obtain z-tester multiple. It uses NORMSINV excel function.

p-value for every likely test conditions is calculated and compared to every alpha level to decide whether to reject null hypothesis. The various inputs include;

0.4 – population proportion for hypothesis

0.4 – sample proportion

The right number of the situation is then put. Following calculations include;

0.09 as standard error

0.54 as z-test statistic

The result includes;

For the provided alpha level :

H0: P => 0.36

 P-value = 0.72

One tailed

The H0 for this situation should be 0.1

H0: P < = 0.35                      - no rejection

P-value = 0.28 -                      one tailed

 

H0: P = 0.35                        no rejection

P-value = 0.57                    two tailed

 

The H0 for the situation must be  0.1 not rejected , 0.5 not rejected.

552 Words  2 Pages
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