What is the equation of the line that passes through the point (-3,-4) and has a slope of 4?

A. A researcher reads that in the population of US students, the average income of students is $1500. In a random sample of 300

MrRa [10]

Answer:

A. Value of t = -1.7974 .

B. If the t-score that was calculated was between 1.7 and 1.9, we will not be able to reject the null hypothesis.

C. No, we can't reject the null hypothesis with 99% confidence.

Step-by-step explanation:

We are given the population mean income of students, $\mu$ = $1500.

Let, Null Hypothesis, $H_0$ : $\mu$ = $1500

Alternate Hypothesis, $H_1$ : $\mu$ < $1500

A. The one-tailed t statistics used here is ;

$\frac{xbar - \mu}{\frac{s}{\sqrt{n} } }$ follows $t_n_-_1$ where, xbar = sample mean = $1445

s = sample standard deviation = $530

n = sample size = 300

Test Statistics = $\frac{1445 - 1500}{\frac{530}{\sqrt{300} } }$ follows $t_2_9_9$

Value of t = -1.7974

<em>Now, at 5% level of significance the t table will give a critical value of -1.9696. Since our test statistics is greater than critical value as -1.7974 > -1.9696 so we have insufficient evidence to reject null hypothesis and conclude that population mean is $1500 and sample mean is less than the population mean</em>.

B. If the t-score that was calculated above was between 1.7 and 1.9 then also we haven't rejected the null hypothesis as then also our test statistics doesn't lie in the rejection region as 1.7 to 1.9 is greater than -1.9696. And we conclude that sample and the population mean are not same.

C. <em>99% Confidence Interval for </em> $\mu$ = $[xbar - 1.9696*\frac{s}{\sqrt{n} } , xbar + 1.9696*\frac{s}{\sqrt{n} }]$