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dlinn [17]
3 years ago
12

paid after the grace period, on average, more than 4 times in 2018, what are the null and alternative hypotheses? Select the cor

rect answer below: H0: μ=4; Ha: μ>4 H0: μ>4; Ha: μ=4 H0: μ=4; Ha: μ<4 H0: μ=4; Ha: μ≠4
Mathematics
1 answer:
Mila [183]3 years ago
4 0

Answer:

<em>H</em>₀: <em>μ</em> = 4 vs. <em>H ₐ</em>: <em>μ </em>> 4

Step-by-step explanation:

A null hypothesis is a sort of hypothesis used in statistics that intends that no statistical significance exists in a set of given observations.  

It is a hypothesis of no difference.

It is typically the hypothesis a scientist or experimenter will attempt to refute or discard. It is denoted by H₀.

Whereas, the alternate hypothesis is the contradicting statement to the null hypothesis.

The alternate hypothesis describes direction of the hypothesis test, i.e. if the test is left tailed, right tailed or two tailed.

It is also known as the research hypothesis and is denoted by H ₐ.

In this case we need to test whether the amount is paid after the grace period, on average, more than 4 times in 2018.

The hypothesis can be defined as follows:

<em>H</em>₀: <em>μ</em> = 4 vs. <em>H ₐ</em>: <em>μ </em>> 4

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Step-by-step explanation:

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Andreas93 [3]

Answer:

0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151

0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233

Step-by-step explanation:

Information given

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n= 250 represent the sample size

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In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=\pm 1.645

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151

0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233

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3 years ago
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