01:960:401 Lecture Notes - Lecture 16: Statistical Hypothesis Testing, Standard Deviation, Random Variable

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lavendertortoise233

15 May 2018

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Chapter 10

Section 9: Comparing proportions from 2 populations.

Question: How do we make inferences about

and

Answer: We introduce a new random variable

ˆˆ

pp−

, where

ˆx

, and

ˆx

where

is the number of successes in sample 1 and

is the number of trials in a binomial experiment

from population 1, and

is the number of successes in sample 2 and

is the number of trials

in a binomial experiment from population 2. It turns out that the distribution of

ˆˆ

pp−

has a

mean of

pp−

and a standard deviation of

1 1 2 2

p q p q

Large sample hypothesis test of

pp−

, based on

ˆˆ

pp−

The null and alternative hypothesis can take on one of the following form:

0 1 2

:H p p=

0pp−=

1 1 2

:H p p

0pp−

(left-tailed test)

0 1 2

:H p p=

0pp−=

1 1 2

:H p p

0pp−

(right-tailed test)

0 1 2

:H p p=

0pp−=

1 1 2

:H p p

0pp−

(two-tailed test)

We assume the samples are independent and

are all greater than or

equal to 10.

Recall when doing hypothesis testing we assume

is true. When we did a hypothesis test for

one proportion we used the hypothesized

when calculating



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Document Summary

Answer: we introduce a new random variable p. 2 where 1x is the number of successes in sample 1 and. 1n is the number of trials in a binomial experiment from population 1, and. 2x is the number of successes in sample 2 and. 2n is the number of trials in a binomial experiment from population 2. It turns out that the distribution of p. 1 and a standard deviation of p q. The null and alternative hypothesis can take on one of the following form: We assume the samples are independent and 1 1 n p , 1 1 n q , equal to 10. 2 2 n q are all greater than or. When we did a hypothesis test for one proportion we used the hypothesized p when calculating. Hence we must calculate a pooled estimate for p from information from both samples. The pooled estimate for p, denoted p , is p x.

Related Questions

(1 point) A random sample of 130 observations is selected from a binomial population with unknown probability of success p. The computer value of is 0.64.

(1) Test : : p = 0.65 against : p > 0.65. Use α= 0.05.

test statistic z =

critical z score

The final conclusion is

⭕ A. We can reject the null hypothesis that p = 0.65 and accept that p is > 0.65.

⭕ B. There is not sufficient evidence to reject the null hypothesis that p = 0.65.

(2) Test : p = 0.5 against :p < 0.65. Use α = 0.01.

test statistic z =

critical z score

The final conclusion is

⭕ A. We can reject the null hypothesis that p = 0.5 and accept that p is < 0.5.

⭕ B. There is not sufficient evidence to reject the null hypothesis that p = 0.5.

(3) Test : p = 0.55 against : p ≠ 0.55. Use α = 0.01.

test statistic z =

positive critical z score

oliver202166

A salesperson makes four calls per year. A sample of 100 days given the following frequencies of sales volumes.

Number of Sales	Observed Frequency (days)
0	30
1	32
2	25
3	10
4	3
TOTAL	100

Records show sales are made to 30% of all sales calls. Assuming independent sales calls, the number of sales per day should follow a binomial distribution. The Binomial distribution is represented by:

For this exercise, assume that the population has a binomial probability distribution with n=4, p=0.30 and x= 0, 1, 2, 3 and 4.

(a) Compute the expected frequencies for x=0, 1, 2, 3 and 4 by using the binomial probability function. Combine categories if necessary to satisfy the requirement that the expected frequency is five or more for all categories.

(b) Use the goodness of fit test to determine whether the assumption of a binomial probability distribution should be rejected. Because no parameters of the Binomial probability distribution were estimated from the sample data, the degrees of freedom are k-1 .

whitebear607

(1 point) Independent random samples, each containing 900 observations, were selected from two binomial populations. The samples from populations 1 and 2 produced 632 and 665 successes, respectively.

(a) Test against ≠ 0 . Use α = 0.07