IM having a bad day help pleaseeeee

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> $\mu$ = $[xbar - 1.9696*\frac{s}{\sqrt{n} } , xbar + 1.9696*\frac{s}{\sqrt{n} }]$

= $[1445 - 1.9696*\frac{530}{\sqrt{300} } , 1445 + 1.9696*\frac{530}{\sqrt{300} }]$

= [ 1384.731 , 1505.269 ]

Since, $1500 lies in this 99% confidence interval so still we can't reject null hypothesis.

7 0

3 years ago

The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead samp

Lesechka [4]

Answer:

e. $1.8\times 10^{-6}m^3$

Step-by-step explanation:

It is given that,

The density of intergalactic space material is $2.5 \times 10^{-27}$ kg per cubic meter.

And the volume of intergalactic space material is $8.0\times 10^{24} m^3$

So the mass of that much intergalactic space material is,

$m= \rho\times v$ , where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

$m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}$

$m=20\times 10^{(-27+24)}=20\times 10^{-3} kg$

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is $20 \times 10^{-3}=2 \times 10^{-2}kg$