Question

A development economist is interested in whether average years of schooling of girls and boys are the same in a certain developing country. A random sample of 250 girls yields a mean of 5.1 years, and

asked Dec 9, 2020 94.4k views

A development economist is interested in whether average years of schooling of girls and boys are the same in a certain developing country. A random sample of 250 girls yields a mean of 5.1 years, and a standard deviation of 1.2 years. An independent random sample of 280 boys yields a mean of 6.3 years, and a standard deviation of 1.7 years.Test the null hypothesis that mean years of schooling is the same in the populations of girls (X population) and boys (Y population), against the alternative hypothesis that the population means are different. Use a 5% level of significance.1. What is the value of the lower end of the confidence interval?a. – 1.4086

b. – 1.4485
c. – 1.3715
d. – 1.40442. What is the value of the upper end of the confidence interval?a. – 1.2085
b. – 0.9956
c. – 0.9515
d. – 0.99143. What kind of distribution ( central limit theorem or T) and why?

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Armonge · Answer 1 · 2020-12-13T11:20:33+0000

Answer:

So on this case the 95% confidence interval would be given by
$-1.4485 \leq \mu_G -\mu_B \leq -0.9515$

Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.

1. What is the value of the lower end of the confidence interval?

b. – 1.4485

2. What is the value of the upper end of the confidence interval?

c. – 0.9515

What kind of distribution ( central limit theorem or T) and why?

We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom of the t distribution increases we have a normal distribution.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =5.1$ represent the sample mean 1 (girls)

$\bar X_2 =6.3$ represent the sample mean 2 (boys)

n1=250 represent the sample 1 size

n2=280 represent the sample 2 size

s_1 =1.2 sample standard deviation for sample 1

s_2 =1.7 sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}$ (1)

The point of estimate for
$\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =5.1-6.3=-1.2$

In order to calculate the critical value
$t_(\alpha/2)$ we need to find first the degrees of freedom, given by:

df=n_1 +n_2 -1=250+280-2=528

Since the Confidence is 0.95 or 95%, the value of
$\alpha=0.05$ and
$\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,528)".And we see that
$t_(\alpha/2)=1.964$

The standard error is given by the following formula:

$SE=\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}$

And replacing we have:

$SE=\sqrt{(1.2^2)/(250)+(1.7^2)/(280)}=0.127$

Now we have everything in order to replace into formula (1):

$-1.2-1.96\sqrt{(1.2^2)/(250)+(1.7^2)/(280)}=-1.4485$

$-1.2+1.96\sqrt{(1.2^2)/(250)+(1.7^2)/(280)}=-0.9515$