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alex41 [277]
3 years ago
15

For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test

statistics.(a) z0 = 2.05 (b) z0 = −1.84 (c) z0 = 0.4
Mathematics
1 answer:
Basile [38]3 years ago
3 0

Answer:

a) p_v =P(Z>2.05)=1-P(z

b) p_v =P(Z>-1.84)=1-P(z

c) p_v =P(Z>0.4)=1-P(z

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: \mu=10

Alternative hypothesis: \mu >10

If the random variable is distributed like this: X \sim N(\mu,\sigma)

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

z_o=\frac{\bar X -\mu}{\sigma}

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

p_v =P(Z>2.05)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

p_v =P(Z>-1.84)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

p_v =P(Z>0.4)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say \alpha=0.05. For part a the p_v so then we can reject the null hypothesis at this significance level.

For part b the p_v>\alpha so then we FAIL to reject the null hypothesis at this significance level.

For part c the p_v>\alpha so again we FAIL to reject the null hypothesis at this significance level.

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Step-by-step explanation:

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4 years ago
Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1
svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{9.3 - 8}{0.4919}

Z = 2.64

Z = 2.64 has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

n = 6 means that s = \frac{1.1}{\sqrt{6}} = 0.4491

This probability is 1 subtracted by the pvalue of Z when X = 9. So

Z = \frac{X - \mu}{s}

Z = \frac{9 - 8}{0.4491}

Z = 2.23

Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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