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True [87]
3 years ago
6

Help me please and can i have a friend

Mathematics
2 answers:
soldier1979 [14.2K]3 years ago
5 0

Answer:

7440 yards and sure

Step-by-step explanation:

1760*4=7040

7040+400=7440

stellarik [79]3 years ago
4 0

Answer:

7440 yards

Step-by-step explanation:

If 1 mile is 1760 yards, to find 4 miles, you multiply 4 by 1760.

1760*4=7040

7040 yards+ 400 yards = 7440 yards

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Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp
exis [7]

Answer:

H_0: \, \mu = 2.

H_1:\, \mu > 2.

Test statistics: z \approx 2.582.

Critical value: z_{1 - 0.025} \approx 1.960.

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean \mu is greater than 2. This claim should be reflected in the alternative hypothesis:

H_1:\, \mu > 2.

The corresponding null hypothesis would be:

H_0:\, \mu = 2.

In this setup, the null hypothesis H_0:\, \mu = 2 suggests that \mu_0 = 2 should be the true population mean of GPA.

However, the alternative hypothesis H_1:\, \mu > 2 does not agree; this hypothesis suggests that the real population mean should be greater than \mu_0= 2.

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed \mu_0 = 2 (i.e., the null hypothesis is true.) How likely would the sample (sample mean \overline{X} = 2.02 with sample standard deviation s = 0.06) be observed in this hypothetical population?

Let \sigma denote the population standard deviation.

Given the large sample size n = 60, the population standard deviation should be approximately equal to that of the sample:

\sigma \approx s = 0.06.

Also because of the large sample size, the central limit theorem implies that Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} should be close to a standard normal random variable. Use a Z-test.

Given the observation of \overline{X} = 2.02 with sample standard deviation s = 0.06:

\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}.

Because the alternative hypothesis suggests that the population mean is greater than \mu_0 = 2, the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed z-test. The question requested a significant level of 0.025. Therefore, the critical value z_{1 - 0.025} should ensure that P( Z > z_{1 - 0.025}) = 0.025.

Look up an inverse Z table. The z_{1 - 0.025} that meets this requirement is z_{1 - 0.025} \approx 1.960.

The z-value observed from the sample is z_\text{observed}\approx 2.582, which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this 0.025 confidence level in favor of the alternative hypothesis.

3 0
3 years ago
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