Statistics Pape Example: One-Sample Hypothesis Test

For the one-sample hypothesis test, the null and alternative hypotheses are:

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Null hypothesis Hnull: The Republican candidate George W. Bush will not win the state by winning more than 50% (0.5) of the votes at the 0.10 significance level (p = 0.5)

Alternative hypothesis Halt: The Republican candidate George W. Bush will win the state by winning more than 50% (0.5) of the votes at the 0.10 significance level (p > 0.5)

Where,

p is the probability

Testing the hypothesis:

In order to test the hypothesis above we will utilize a one-sample z test.

For the z test, the test statistic is:

z= p-pp 1-p/ nWhere,

p is the sample proportion for George Bush (Calculated below)

p is the probability = 0.5

n is the sample size = 765

We can compute the sample proportion (p) as:

p= xnWhere,

x is the number of votes cast for George Bush = 407

n is the total sample size = 765

Therefore, the sample proportion is equal to:

p= 407765=0.532Using this value, we can compute the value of the test statistic as:

z= 0.532-0.50.5 1-0.5/ 765=1.77Using this tests statistic, we can determine the p-value

P-value = P (z > 1.77) = 1 p (z < 1.77)

And from the z-table,

P (z < 1.77) = 0.9616

Therefore,

The p-value = 1 0.9616

= 0.0384

From the results obtained above it is clear that George will win the state by winning more than 50% (0.5) of the votes at the 0.10 significance level (p > 0.5). The main reason for this observation is that Z = 1.77, p = 0.0384. This implies that the z tests results are significant at the 0.1 significance level, therefore, we reject the null hypothesis.

Case Study SpeedX

Hypothesis:

For the one-sample hypothesis test, the null and alternative hypotheses are:

Null hypothesis Hnull: The mean payment period is equal to 22 days at the 0.10 significance level (u = 0.5)

Alternative hypothesis Halt: The mean payment period is less than 22 days at the 0.10 significance level (u < 22)

Where,

u is the population mean

Testing the hypothesis:

In order to test the hypothesis above we will utilize a one-sample z test.

For the z test, the test statistic is:

z= x- us nWhere,

x is the sample proportion

u is the population mean = 22

s is the standard deviation = 6

n is the sample size = 220

We can compute the sample proportion (x) as:

x= xnWhere,

x is the total number of hours = 4759

n is the total sample size = 220

Therefore, the sample proportion is equal to:

x= 4759220=21.63Using this value, we can compute the value of the test statistic as:

z= 21.63-226 / 220=-0.91Using this tests statistic, we can determine the p-value

P-value = P (z < - 0.91)

And from the z-table,

P (z < - 0.91) = 01814

From the results obtained above it is clear that we can fail to reject the null hypothesis. The main reason for this is that the p-value (0.1814) is greater than alpha (0.1) Z = - 0.91, p = 0.1814. This implies that the z tests results are not significant at the 0.1 significance level; therefore, we fail to reject the null hypothesis.