Please help!! if you could show work too that would be amazing :))

Until recently a number of professions were prohibited from advertising. In 1977, the U.S. Supreme Court ruled that prohibiting

umka2103 [35]

Answer:

the two groups differed in their attitudes toward advertising, significantly

Step-by-step explanation:

given that until recently a number of professions were prohibited from advertising. In 1977, the U.S. Supreme Court ruled that prohibiting doctors and lawyers from advertising violated their right to free speech. The paper "Should Dentists Advertise?" Journal of Advertising Research, June 1982) compared the attitudes of 101 consumers and 124 dentists to the questions "I favor the use of advertising by dentists to attract new patients".

Hypotheses would be

H-0: the groups do not differ

H-a: The two groups differ significantly

(two tailed chi square test)

Observed

Group Strongly agree Agree Neutral Disagree Strongly disagree Total

Consumers 34 47 9 6 5 101

Dentists 9 18 23 28 46 124

Expected is calculated as row total * column total / grand total for each cell

chi square = (obs - exp)^2/Exp for each cell added

chi square = 79.2716

rxc 5x2

df = 4x1 =4

p value <0.00001

Since p < alpha say 0.01, we reject null hypothesis.

the two groups differed in their attitudes toward advertising, significantly

3 0

3 years ago

Solve the inequality.<br> I need help please.<br> -11.6>6.7+4.3+m

satela [25.4K]

First, combine like terms:

(6.7+4.3= 11)

-11.6>11+m
isolate m on one side
-11.6-11>m
=
m<-22.4

6 0

3 years ago

The standard deviation for actuary salaries has a standard deviation of $36,730 . You collect a simple random sample of n=36 sal

Yanka [14]

Using the z-distribution, it is found that the lower limit of the 95% confidence interval is of $99,002.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.