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Natali5045456 [20]
1 year ago
12

Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test

statistic, critical value(s) , P-value , and state the final conclusion.Test the claim that for the population of female college students, the mean weight is given by μ = 132 lb. Sample data are summarized as n = 20, overbar(x) = 137 lb, and s = 14.2 lb. Use a significance level of α = 0.1.
Mathematics
1 answer:
BaLLatris [955]1 year ago
7 0

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\  \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}

- Applying the formula, we have:

\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\  \\ t=\frac{5}{3.1752} \\  \\ t\approx1.5747 \end{gathered}

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

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Let X denote the number of flaws along a 100-m reel of magnetic tape (an integer-valued variable). Suppose Zdenote the number of
Lemur [1.5K]

Answer:

a) P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728

b) P(X \leq 30)= P(X

And using the z score we got:

P(X

c) P(X

And if we use the continuity correction we got:

P(X

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(25,5)  

Where \mu=25 and \sigma=5

Part a

For this case we want to find this probability:

P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

And if we use the z score given by:

z =\frac{x-\mu}{\sigma}

We got this:

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728

Part b

For this case we want this probability:

P(X \leq 30)

And if we use the continuity correction we got:

P(X \leq 30)= P(X

And using the z score we got:

P(X

Part c

For this case we want this probability:

P(X

And if we use the continuity correction we got:

P(X

3 0
3 years ago
Kris has finished 7/8 of 40 questions. how many were done​
VLD [36.1K]

7/8=0.875

0.875 x 40= 35

Therefore, Kris has finished 35 questions.

Hope this helps.

8 0
2 years ago
Find the area of the regular polygon. Round to the nearest tenth.
mojhsa [17]

Given:

Side length = 12 in

To find:

The area of the regular polygon.

Solution:

Number of sides (n) = 6

Let us find the apothem using formula:

$a=\frac{s}{2 \tan \left(\frac{180^\circ}{n}\right)}

where s is side length and n is number of sides.

$a=\frac{12}{2 \tan \left(\frac{180^\circ}{6}\right)}

$a=\frac{6}{ \tan (30^\circ)}

$a=\frac{6}{ \frac{1}{\sqrt{3} }}

$a=6\sqrt{3}

Area of the regular polygon:

$A=\frac{1}{2}(\text { Perimeter })(\text { apothem })$

$A=\frac{1}{2}(6 \times 12)(6\sqrt{3} )

$A=\frac{1}{2}(72)(6\sqrt{3} )

A=216 \sqrt{3}

A=374.1 in²

The area of the regular polygon is 374.1 in².

4 0
3 years ago
Please help me!! (Look at the photo)
maw [93]

Answer:

310in

Step-by-step explanation:

7×10=70

70×2=140(there are two that are the same length)

5×10=50

50×2=100(there are two that are the same length)

5×7=35

35×2=70(there are two that are the same length)

70+100=170

170+140=310

please let me know if it's correct

4 0
3 years ago
HELP ME PLZ BRAINLIST 4 U
mylen [45]

Answer:

G

Step-by-step explanation:

I think its G because i counted the numbers...and i pretty sure its not a Negative so yah

8 0
3 years ago
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