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Yakvenalex [24]
3 years ago
9

A professor of statistics wants to test that the average amount of money a typical college student spends per day during spring

break is over $70. Based upon previous research, the population standard deviation is estimated to be $17.32. The professor surveys 35 students and finds that the mean spending is $72.43. Which of the following statements is most accurate? Select one: a. reject the null hypothesis at ? = 0.01 b. reject the null hypothesis at ? = 0.10 c. fail to reject the null hypothesis at ? ? 0.10 d. reject the null hypothesis at ? = 0.05
Mathematics
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

Option c: Fail to reject the null hypothesis at α = 0.10

Step-by-step explanation:

We are given;

Population mean; μ = 70

population standard deviation; σ = 17.32

Sample mean; x¯ = 72.43

Sample size; n = 35

Thus;

Null hypothesis; H0: μ = 70

Alternative hypothesis; Ha: μ > 70

Z-score formula is;

z = (x¯ - μ)/(σ/√n)

z = (72.43 - 70)/(17.32/√35)

z = 2.43/2.9276

z = 0.83

From online p-value from z-score calculator attached, using z = 0.83; significance level = 0.05 and one tailed hypothesis, we have;

P-value ≈ 0.2033

It's more than the significance value, so we will fail to reject the null hypothesis at α = 0.05

Similarly, From online p-value from z-score calculator attached, using z = 0.83; significance level = 0.01 and one tailed hypothesis, we have;

P-value ≈ 0.2033

It's more than the significance value, so we will fail to reject the null hypothesis at α = 0.01

Similarly, From online p-value from z-score calculator attached, using z = 0.83; significance level = 0.10 and one tailed hypothesis, we have;

P-value ≈ 0.2033

It's more than the significance value, so we will fail to reject the null hypothesis at α = 0.10

From the 3 significance values, the correct option is option c. fail to reject the null hypothesis at α = 0.10

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3 years ago
The speed at which cars travel on the highway has a normal distribution with a mean of 60 km/h and a standard deviation of 5 km/
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The z-score of the speed value gives the measure of dispersion of the from

the mean observed speed.

The probability that the speed of a car is between 63 km/h and 75 km/h is

<u>0.273</u>.

The given parameters are;

The mean of the speed of cars on the highway, \overline x = 60 km/h

The standard deviation of the cars on the highway, σ = 5 km/h

Required:

The probability that the speed of a car is between 63 km/h and 75 km/h

Solution;

The z-score for a speed of 63 km/h is given as follows;

Z=\dfrac{x-\bar x }{\sigma }

Which gives;

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From the z-score table, we have;

P(x < 63) = 0.7257

The z-score for a speed of 75 km/h is given as follows;

Z=\dfrac{75-60 }{5 } = 3

Which gives, P(x < 75) = 0.9987

The probability that the speed of a car is between 63 km/h and 75 km/h is therefore;

P(63 < x < 75) = P(x < 75) - P(x < 63) = 0.9987 - 0.7257 = 0.273

The probability that the speed of a car is between 63 km/h and 75 km/h is

<u>0.273</u>.

Learn more here:

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