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faust18 [17]
3 years ago
13

To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo

rtive of its cause. They ask a sample of 5 men and 5 women to rate the importance of their cause on a scale from 1 (not important at all) to 7 (very important). The ratings for men were M1 = 6.2. The ratings for women were M2 = 5.3. If the estimated standard error for the difference is equal to 0.25, then consider the following. (a) Find the confidence limits at an 80% CI for these two independent samples. (Round your answers to two decimal places.)
Mathematics
1 answer:
Naya [18.7K]3 years ago
4 0

Answer:

(0.55525, 1.24475)

Step-by-step explanation:

Given that to save money, a local charity organization wants to target its mailing requests for donations to individuals who are most supportive of its cause.

                     Men    women

Sample size      5        5

Mean rate        6.2     5.3

Mean diffference = 6.2-5.3=0.9

Std error for difference = 0.25 (given)

Since we have two groups and also sample sizes are very small we can use t significant value for finding out confidence interval

df = 5+5-2 =8

t critical value for 80% two tailed = 1.397

Conf interval 80%

=(0.9-1.397*0.25, 0.9+1.397*0.25)\\= (0.55525, 1.24475)

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Answer:

a) \bar X = 369.62

b) Median=175

c) Mode =450

With a frequency of 4

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<u>e)</u>s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

Replacing we got:

\bar X = 369.62

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Median=175

Part c

The mode is the most repeated value in the sample and for this case is:

Mode =450

With a frequency of 4

Part d

The midrange for this case is defined as:

MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

Part e

For this case we can calculate the deviation given by:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

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