Question

lots of parts please help!

Researchers wondered if there was a difference between male and females in regard to some common annoyances They asked a random sample of males and femalesthe following question "Are you arvioyed by people who repeatedly check the mobile phones while having an in person conversation Among the 513 males surveyed. 194 responded "Yesamong the 551 females surveyed. 212 responded "Yes" Does the evidence suggest a Highet proportion of females we annoyed by this behavior? Complete parts (1) through ( below (1) Determine the sample proportion for each sample The proportom of the females and males who took the survey who are noyed by the behavior in question are and respectively Round to four decimal places as needed) () Explain why this study can be analyzed using the methods for conducting a hypothesis est regarding two independent proportion Select all that DA The data come from a population that is normally due D. The sample se less than the populations for each and DC. The samples are independent D. The samples we dependent OL (-) 10 dny (...) DF. The sample size is more than 5% of the population site for each sample Ic) What are the wild temative hypotheses? Let , represent the population proportion of nes who are annoyed by the behavior in ton and present the population proportion of males who are annoyed by the Closiect your we

Answer 1

a)

Answer:

Proportion of female = 0.4011

Proportion of male = 0.3573

Explanation:

221 Proportion of female, P1 = = 0.4011 551

Proportion of male, P1 194 543 - 0.3573

b)

Answer:

Correct option: B, C, E

Explanation:

The necessary conditions to perform a hypothesis test is shown below,

1) The normality condition:  

$n\widehat{p}_1(1-\widehat{p}_1)\geq 10\text{ and }n\widehat{p}_2(1-\widehat{p}_2)\geq 10$

2) The independence condition: The sample size is 5% or less of the population

3) Simple random sample: The samples are collected by a simple random sampling method.

c)

Answer:

$H_0:p_1=p_2$

Explanation:

The null hypothesis is defined as the proportion of females and males who are annoyed by this behavior is equal and the alternative hypothesis tests the claim that the proportion of females is higher compared to the proportion of males

d)

Answer:

mean = 0.0438

standard deviation = 0.0293

Ос а

Explanation:

The sample distribution of the difference of sample proportions follows a normal distribution with,

$\text{mean}=\widehat{p}_1- \widehat{p}_2=0.4011-0.3573=0.0438$

$\text{standard deviation}=\sqrt{\overline{p}(1-\overline{p})\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}$

where

$\overline{p}=\frac{x_1+x_2}{n_1+n_2}=\frac{221+194}{551+543}=0.3793$

now,

$\text{standard deviation}=\sqrt{0.3793(1-0.3793)\left ( \frac{1}{551}+\frac{1}{543} \right )}=0.0293$

Since we are performing a right-tailed test, option C is correct.

e)

Answer:

Test statistic = 1.49

P-value = 0.068

Explanation:

Test statistic

The z-statistic is obtained using the following formula,

$z=\frac{\left (\widehat{p}_1-\widehat{p}_2 \right )-0}{\sqrt{\overline{p}(1-\overline{p})\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}$

$z=\frac{0.0438-0}{0.0293}=1.4933\approx 1.49$

P-value

The P-value for the z-statistic is obtained from the z distribution table for z = 1.49

P-value = 0.068

f)

Answer:

Explanation:

If the population proportions are equal, one would expect a sample difference proportion zero

If we do repeated samples, one would expect 68 out of 1000 repetitions of this experiment.

g)

Answer:

B.

Explanation:

Since the p-value = 0.068 is greater than 0.05 at a 5% significance level, the null hypothesis is not rejected.

Hence there is not sufficient evidence to conclude that a higher proportion of females are annoyed by the behavior in question.