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skad [1K]
1 year ago
11

Joe hypothesizes that the students of an elite school will score higher than the general population. He records a sample mean eq

ual to 568 and states the hypothesis as μ = 568 vs μ > 568. What type of test should Joe do?
Mathematics
1 answer:
Andrew [12]1 year ago
7 0

Answer:

The test to be used is the right tailed test.

Step-by-step explanation:

The type of test joe should do would be a right tailed test. This is because;

A right tailed test which we sometimes call an upper test is where the hypothesis statement contains the greater than (>) symbol. This means that, the inequality points to the right. For example, we want to compare the the life of batteries before and after a manufacturing change.

If we want to know if the battery life of maybe 90 hours would be greater than the original, then our hypothesis statements might be:

Null hypothesis: (H0 = 90).

Alternative hypothesis: (H1) > 90.

In the null hypothesis, there are no changes, but in the alternative hypothesis, the battery life in hours has increased.

So, the most important factor here is that the alternative hypothesis (H1) is what determines if we have a right tailed test, not the null hypothesis.

Thus, the test to be used is the right tailed test.

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If candies are sold at 3 pcs for php.2.00.how many candies can melai get if she has php20.00?​
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Which of the following values are in the range of the function graphed below?
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The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin
Free_Kalibri [48]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.03

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03  

= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03  

=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12  

Thus, 8.17 g separates the top 3% of the weights.

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90  

Thus, 7.90 separates the bottom 3% of the weights.

Thus, the correct answer is

Option D) 7.90 g and 8.12 g

7 0
2 years ago
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