Ask question

Ask question

All categories

3 years ago

5

Before the 1936 US presidential election, a popular poll predicted that Alfred Landon would comfortably win against Franklin D.

Roosevelt. The poll randomly sampled people from sources like telephone and car registration records, but the country was in an economic crisis, so many voters without those luxuries were not included in the poll. Voters with lower incomes favored Roosevelt, who ultimately won with 62%, percent of the vote—one of the largest margins of victory ever in a US presidential election. Which type of bias does this scenario best describe?
a. Bias from undercoverage
b. Bias from using voluntary response
c. Response bias
d. Nonresponse bias

1 answer:

Vinvika [58]3 years ago

3 0

Answer:

wrong subject

Step-by-step explanation:

You might be interested in

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

Why lie about my free answers );

dimaraw [331]

I like your sketchers

3 0

3 years ago

Need help ASAP and orange please I’ll give brainiest

Karo-lina-s [1.5K]

Answer:

A

Step-by-step explanation:

Just switch the orders around where it is going from the highest power to the lowest power

6 0

2 years ago

Read 2 more answers

About 62% of the human body is composed of water. Write this percent as a fraction in simplest form.

aleksklad [387]

<span>31/50 by the way I love how some people realize so many people answered this question already and decide they should add something that is totally not necessary.

</span>

7 0

3 years ago

Read 2 more answers

An icicle has a volume of 3769.91 mm if this icicle Drips at a rate of 75 mm/min How long will it take the icicles completely me

kvasek [131]

Answer:

The answer is 3,019.92 cubic millimeter

Step-by-step explanation:

take 375 plus 375 to get 750

then 3769.91 - 750 to get the 3,019.92

3 0

3 years ago

Read 2 more answers