I need helpppp :((((((( 20 marks for this summative

Mathematics

1 answer:

KonstantinChe [14]2 years ago

5 0

Answer:

47v

Step-by-step explanation:

You might be interested in

Tanya spent $8.15 on 5 cans of tuna. She used a coupon that gave her 35 cents off each can. What was the price of a can of tuna

user100 [1]

she spent 1.98 ..............

5 0

2 years ago

53% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a pr

koban [17]

Can you be more specific about what is P(5)? The probability would probably be around five though.

5 0

3 years ago

What is the value of x?

Marrrta [24]

I dont know i just want coins lol
plus i just dont know

3 0

2 years ago

You want to get a pizza the pizza is 14.00 but 50% off. How much will the pizza cost with 8% delivery tax Fees included?

nignag [31]

Answer:

7.56

Step-by-step explanation:

14x.50(the remainder after you subtract the percent from 100) = 7

8% of 7 is 0.56

7+0.56=7.56

4 0

2 years ago

A manufacturer produces crankshafts for an automobile engine. the crankshafts wear after 100,000 miles (0.0001 inch) is of inter

Daniel [21]

Part A:

Significant level:

α = 0.05

Null and alternative hypothesis:

h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{2.78-3}{0.9/\sqrt{15}} \\ \\ = \frac{-0.22}{0.2324} =-0.9467$

P-value:

P(-0.9467) = 0.1719

Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438

Conclusion:

Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.

Part B:

The power of the test is given by:

$\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\ \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\ \\ =0.8117-0.0012=0.8105$

Therefore, the power of the test if μ = 3.25 is 0.8105.

Part C:

The sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:

$1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\ \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\ \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\ \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898$

$\Rightarrow n=(3.8898)^2=15.13$

Therefore, the sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.

8 0

2 years ago