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2 years ago

5

The coldest temperature ever recorded in Nome, Alaska, is -54°F. The average temperature in Nome during the Serum Run was 34° wa

rmer. What was the average temperature in Nome during the Serum Run?

1 answer:

quester [9]2 years ago

7 0

Answer = -54 + 34 = -20 °F
Since it was warmer it means there was an increase in temperature

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I think there are 23 students in the art class.

Step-by-step explanation:

You first subtract 9 from 101. Then, you divide by four to get 23.

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One-hundred students were allowed to re-take an exam for their math course. The probability distribution shows how studying for

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0.912 (C) is the probability that a student who saw an increase in their grade had studied

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A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

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Answer:

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Step-by-step explanation:

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