A) When conducting a hypothesis test with a binomial distribution (sometimes cal

Question

A) When conducting a hypothesis test with a binomial distribution (sometimes called a Binomial Test), there are three ways to calculate the P-value (with additional variations possible). The only exact calculation is to use the binomial probability distribution. The other methods are approximations using the standardized normal distribution (when certain criteria have been achieved). Of these two methods, one can use the sample counts or one can use the sample proportions. Furthermore, it is possible in both of these approximating cases to apply a continuity correction to account for the use of a continuous distribution to approximate a discrete distribution.

This problem introduces the method to obtain an exact P-value using the binomial distribution. Also, this method uses Excel to obtain the answers.

For this demonstration problem, we will test a hypothesis that a population proportion has increased since the last time it was measured. Previously, the population proportion was measured at 16%. For the current analysis, a sample of n=150n=150 randomly chosen subjects was obtained, and 29 of those demonstrated the observation of interest (i.e., a success).

To start, we clearly construct the hypotheses for this problem. Because the researcher suggested the proportion has increased, this would suggest a one-tailed test (as can be seen in the choice of HaHa):  
      Ho:p=0.16Ho:p=0.16
      Ha:p>0.16Ha:p>0.16

Using the binomial distribution, the test statistic from the sample would simply be the sample count (the number of successful observations):
      k=29k=29

The P-value for this scenario would be observing this count or one more extreme. With the alternative hypothesis suggesting that values at or below 16% would be unsurprising, this would suggest that the observed count or larger would constitute the potentially extreme responses. Thus, the P-value would be:
      P(X29p=0.16,n=150)P(X29p=0.16,n=150)

Using Excel, this value can be calculated exactly:

=1-BINOMDIST(28,150,16%,TRUE)

which should return a P-value of 0.1579. Thus, with a traditional significance level of either =0.05=0.05 or =0.01=0.01, this P-value would result in failing to reject the null hypothesis. Thus, there is not enough sample evidence to support the claim that the population proportion has increased.

Note: For a two-tailed hypothesis test, the calculation can become a bit tricky. In this case, it is necessary to take the counts at and above the sample count, but it is also necessary to determine comparable counts below the hypothesized population count. This would be obtained using the formula

np(knp)np-(k-np)

This value is the count below the hypothesized mean count by the same distance as the sample count was above the mean count. For this demonstration example, this value would be 19. To obtain the P-value from Excel, you would use the following formula:

=BINOMDIST(19,150,16%,TRUE)+1-BINOMDIST(28,150,16%,TRUE)

Though, it is recommended that the two-tailed test only be used with one of the approximation methods as the calculations are less cumbersome.

Exercise Problem
In 1976, only 6% of the students in the city school district were classified as being learning disabled. A school psychologist suspects that the proportion of learning-disabled children has increased dramatically over the years. To demonstrate this point, a random sample of n=200n=200 students is selected. In this sample there are 20 students who have been identified as learning-disabled. You will use this information to determine if the sample indicates a change in the proportion of learning-disabled students at a 0.01 level of significance.

What is the hypothesized (assumed constant) population proportion for this test?
p=p=  
(Report answer as a decimal accurate to 2 decimal places. Do not report using the percent symbol.)

Based on the researcher's understanding of the situation, how many tails would this hypothesis test have?

one-tailed test

two-tailed test

Choose the correct pair of hypotheses for this situation:

(A)

(B)

(C)

(D)

(E)

(F)

The test statistic for this analysis is the sample count (i.e., the number of observed successes). What is this value?
k=k=

With these hypotheses, the p-value for this test is (assuming HoHo is true) the probability of observing...

at most 20 learning-disabled students

at least 20 learning-disabled students

more than 20 learning-disabled students

at least 12 learning-disabled students

You are now ready to calculate the P-value for this sample. Be sure to use the (cumulative) binomial distribution to obtain an exact P-value. (Do not use the normal distribution as an approximation for the binomial distribution for this particular problem.)
P-value =  
(Report answer as a decimal accurate to 4 decimal places.)

This P-value (and test statistic) leads to a decision to...

reject the null

accept the null

fail to reject the null

reject the alternative

As such, the final conclusion is that...

There is sufficient evidence to warrant rejection of the claim that the proportion of learning-disabled students has increased.

There is not sufficient evidence to warrant rejection of the claim that the proportion of learning-disabled students has increased.

The sample data support the claim that the proportion of learning-disabled students has increased.

There is not sufficient sample evidence to support the claim that the proportion of learning-disabled students has increased.

B) Test the claim that the proportion of people who own cats is smaller than 80% at the 0.05 significance level.

The null and alternative hypothesis would be:

H0:0.8H0:0.8
H1:<0.8H1:<0.8

H0:p0.8H0:p0.8
H1:p<0.8H1:p<0.8

H0:=0.8H0:=0.8
H1:0.8H1:0.8

H0:0.8H0:0.8
H1:>0.8H1:>0.8

H0:p=0.8H0:p=0.8
H1:p0.8H1:p0.8

H0:p0.8H0:p0.8
H1:p>0.8H1:p>0.8

The test is:

right-tailed

two-tailed

left-tailed

Based on a sample of 200 people, 77% owned cats

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Based on this we:

Fail to reject the null hypothesis

Reject the null hypothesis

(A) (B) (C) H0:p=0.06H0:p=0.06
Ha:p<0.06Ha:p<0.06 H0:p=0.06H0:p=0.06
Ha:p0.06Ha:p0.06 H0:p=0.06H0:p=0.06
Ha:p>0.06Ha:p>0.06 (D) (E) (F) H0:p=0.1H0:p=0.1
Ha:p<0.1Ha:p<0.1 H0:p=0.1H0:p=0.1
Ha:p0.1Ha:p0.1 H0:p=0.1H0:p=0.1
Ha:p>0.1Ha:p>0.1

Explanation / Answer

Solution:-

B)

p = 0.77, n = 200

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.80

Alternative hypothesis: P < 0.80

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0283

z = (p - P) /

z = - 1.06

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -1.06. We use the Normal Distribution Calculator to find P(z < -1.06) = 0.1446

Interpret results. Since the P-value (0.1446) is greater than the significance level (0.05), we have to accept the null hypothesis.

Fail to reject the null hypothesis.

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