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Lostsunrise [7]
3 years ago
6

In a survey, of the voters support a particular referendum. If 40 voters are chosen at random, find the mean and variance for th

e number of voters who support the referendum.
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
3 0

This question is incomplete, the complete question is;

In a survey, 55% of the voters support a particular referendum. If 40 voters are chosen at random,

find the mean and variance for the number of voters who support the referendum.

Answer:

a) The Mean is 22

b) Variance is 9.9

Step-by-step explanation:

Given that;

55% of the voters support a particular referendum p = 0.55

q = 1 - p = 1 - 0.55 = 0.45

sample size n = 40

a)

Mean = sample size n × p

Mean = 40 × 0.55

Mean = 22

Therefore the Mean is 22

b)

Variance = n × p × q

so we substitute

Variance = 40 × 0.55 × 0.45

Variance = 9.9

Therefore the Variance is 9.9

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Answer:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

df = n_1 +n_2 -2 = 10+15-2= 23

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Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

\bar X_1 = 1085 sample mean for group 1

\bar X_2 = 1034 sample mean for group 2

n_1 = 10 sample size for group 1

n_2 = 15 sample size for group 2

s_1 = 52 sample deviation for group 1

s_2 = 61 sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: \mu_1= \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

And the statistic is given by:

t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

And replacing we got:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 10+15-2= 23

And the p value would be:

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

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