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AleksandrR [38]
2 years ago
9

Cameron wondered if the average score on a final exam was different between those who texted on a regular basis during the lectu

res for a particular class and those that did not text at all during the lectures. Which of the following statements is correct?
a. A one-tailed test should be performed since the alternative hypothesis states that the parameter is less than the hypothesized value.
b. A one-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.
c. A two-tailed test should be performed since the alternative hypothesis states that the parameter s no e ua to he ypo es e alue.
d. A two-tailed test should be performed since the alternative hypothesis states that the parameter is less than the hypothesized value.
e. A two-tailed test should be performed since the null hypothesis states that the parameter is not equal to the hypothesized value.
Mathematics
1 answer:
Margarita [4]2 years ago
3 0

Answer:

c. A two-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

Step-by-step explanation:

Let p1 be the average score on a final exam who texted on a regular basis during the lectures for a particular class

And p2 be the average score on a final exam who did not texted at all during the lectures for a particular class

According to the Cameron's point of interest, null and alternative hypotheses are:

H_{0}: p1 = p2

H_{a}: p1 ≠ p2

Two tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

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Increasing

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Decreasing

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1<x<4

Step-by-step explanation:

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Given ΔMNO, find the measure of ∠LMN.
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Answer:

  104°

Step-by-step explanation:

If segments NO and NM are congruent, then angles NMO and NOM are congruent. So, their supplements, angles NML and NOP are congruent. That is ...

  ∠NML ≅ ∠NOP = 104°

  ∠NML = 104°

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56/7 = 8 per hat

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An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

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 normally distributed samples are 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.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0
2 years ago
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