Question

1. Independent random samples of n₁ = 200 and n₂ = 200 observations were randomly selected from binomial populations 1 and 2, respectively. Sample 1 had 116 successes, and sample 2 had 122 successes.

a) Calculate the standard error of the difference in the two sample proportions, (p̂₁ − p̂₂). Make sure to use the pooled estimate for the common value of p. (Round your answer to four decimal places.)

b) Critical value approach: Find the rejection region when α = 0.01. (Round your answer to two decimal places. If the test is one-tailed, enter NONE for the unused region.)

z <

z >

2. The meat department of a local supermarket chain packages ground beef in trays of two sizes. The smaller tray is intended to hold 1 kilogram (kg) of meat. A random sample of 30 packages in the smaller meat tray produced weight measurements with an average of 1.01 kg and a standard deviation of 20 grams.

p-value =

Answer 1

using excel>addins >ph stat>two sample test>z test for two proportion

we have

/

Z Test for Differences in Two Proportions
Data
Hypothesized Difference	0
Level of Significance	0.01
Group 1
Number of Items of Interest	116
Sample Size	200
Group 2
Number of Items of Interest	122
Sample Size	200
Std. Error of the Diff. between two Proportions	0.0491
Intermediate Calculations
Group 1 Proportion	0.58
Group 2 Proportion	0.61
Difference in Two Proportions	-0.03
Average Proportion	0.5950
Z Test Statistic	-0.6111
Upper-Tail Test
Upper Critical Value	2.3263
p-Value	0.7294
Do not reject the null hypothesis

a ) the standard error of the difference in the two sample proportions, (p̂1 − p̂2). = 0.0491

b ) z >2.33

Ans 2 ) using excel>addins >phstat>one sample test>one sample t

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	1
Level of Significance	0.05
Sample Size	30
Sample Mean	1.01
Sample Standard Deviation	20
Intermediate Calculations
Standard Error of the Mean	3.6515
Degrees of Freedom	29
t Test Statistic	0.0027
Two-Tail Test
Lower Critical Value	-2.0452
Upper Critical Value	2.0452
p-Value	0.9978
Do not reject the null hypothesis

p value is 0.9978