Mathematics: On average, Americans have lived in 4 places by the time they are 18 years old. Is this average a different number for college students? The 50 randomly selected college students who answered the survey question had lived in an average of 3.79 places by the time they were 18 years old. The standard deviation for the survey group was 1.2. What can be concluded at the α = 0.10 level of significance?

Step-by-step explanation:2+1+5=878.40/8=9.80B=5x9.80=49 C=1x9.80=9.80So B-C=39.20 more...

On average, Americans have lived in 4 places by

marioagundis0998

14.02.2022

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the average is a different number for college students

Step-by-step explanation:

From the question we are told that

The sample size is n = 50

The population mean is $\mu = 4$

The sample mean is $\= x = 3.79$

The standard deviation is $\sigma = 1.2$

The level of significance is $\alpha = 0.10$

The null hypothesis is $H_o : \mu = 4$

The alternative hypothesis is $H_a : \mu \ne 4$

Generally the test statistics is mathematically represented as

$z = \frac{ \= x - \mu }{ \frac{ \sigma }{ \sqrt{n}} }$

=> $z = \frac{ 3.79 - 4 }{ \frac{1.2}{ \sqrt{50}} }$

=> z = -1.237

From the z table the area under the normal curve to the left corresponding to -1.237 is

P(Z < -1.237) = 0.10804

Generally the p-value is mathematically represented as

p-value = P(Z < -1.237) = 2 * 0.10804

=> p-value = 0.21608

Generally from the values obtained we see that

$p-value \alpha$

Hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the average is a different number for college students

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