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posledela
3 years ago
8

University officials say that at least 70% of the voting student population supports a fee increase. If the 95% confidence inter

val estimating the proportion of students supporting the fee increase is [0.75, 0.85], what conclusion can be drawn?
A. 70% is not in the interval, so another sample is needed.
B.70% is not in the interval, so assume it will not be supported.
C.The interval estimate is above 70%, so infer that it will be supported.
D.Since this was not based on population, we cannot make a conclusion.
Mathematics
1 answer:
baherus [9]3 years ago
4 0

Answer:

The interval estimate is above 70%, so infer that it will be supported.

Step-by-step explanation:

The evidence that 95% confidence interval estimating the proportion of students supporting the fee increase is [0.75, 0.85] supports university officials' claim that at least 70% of the voting student population supports a fee increase.

Therefore we can infer that a fee increase will be supported.

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jonny [76]

Answer:

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3 years ago
Please need help thank you
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Answer:

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3 years ago
A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately
Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

z=\frac{\bar x-\mu}{\sigma/\sqrt N}

where

\bar x=mean\; of\;the \;sample

\mu=mean\; established\; in\; H_0

\sigma=standard \; deviation

N = size of the sample.

So,

z=\frac{185-175}{20/\sqrt {10}}=1.5811

\boxed {z=1.5811}

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of  1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of  1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is  

\boxed {p=0.057}

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

<u>Step 1 </u>

Compute \bar x_{critical}

1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}

\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721

So <em>we would make a Type II error if our sample mean is less than 185.3721</em>.  

<u>Step 2</u>

Compute the probability that your sample mean is less than 185.3711  

P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z

So, <em>the probability of making a Type II error is 0.5234 = 52.34% </em>

c)

<em>The power of a hypothesis test is 1 minus the probability of a Type II error</em>. So, the power of the test is

1 - 0.5234 = 0.4766

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6. Fred gave the cashier $20 to pay for 3 large bags of Takis . The cashier gave him $12.56 in change. Each bag of Takis cost th
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Answer:

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Step-by-step explanation:

He paid $20 and got $12.56 back so you subtract

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So his total was 7.44 and he bought 3 bags so divide to find the individual price

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So $2.48

6 0
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