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kondor19780726 [428]

3 years ago

12

Lana finished the race in 13.2 seconds . If 84 other girls ran in the event, approximately how many runners did she beat?

1 answer:

BabaBlast [244]3 years ago

6 0

Complete question :

The 100m dah times in the girl's track meet were normally distributed with a mean of 13 seconds and a standard deviation of 0.3 seconds.

Lana finished the race in 13.2 seconds . If 84 other girls ran in the event, approximately how many runners did she beat?

Answer:

21 runners

Step-by-step explanation:

Mean, μ = 13 seconds

Standard deviation, σ = 0.3

Lana's race time = 13.2

We find the proportion of runners who had race time above 13.2 ;

Proportion of who had race tune above 13.2

P(x > 13.2)

Obtain the Zscore

Zscore = (x - μ) / σ

Z = (13.2 - 13) / 0.3

Zscore = 0.2 / 0.3 = 0.6667

P(Z > 0.6667) = 0.25239 (Z probability calculator)

This is about 0.25239 * 100 = 25.24% = 25% (nearest percent)

Hence, Number of runners Lana beat = 25% of total runners ;

0.25 * 84 = 21

Hence, Lana beat about 21 runners

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Answer:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

$p_v =2*P(t_{30}>2.385)=0.0236$

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

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Step-by-step explanation:

Data given and notation

$\bar X_{insurance}=2.3$ represent the mean for insurance

$\bar X_{No. ins}=1.9$ represent the mean withour insurance

$s_{Insurance}=0.6$ represent the sample standard deviation for the insurance case

$s_{No. Ins}=0.3$ represent the sample standard deviation for the No insurance case

$n_{Insurance}=16$ sample size for insurance

$n_{No. Iss}=16$ sample size for no insurance

t would represent the statistic (variable of interest)

$\alpha=0.01$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis: $\mu_{Insu}=\mu_{No. Ins}$

Alternative hypothesis: $\mu_{Ins} \neq \mu_{No. Ins}$

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

$t=\frac{\bar X_{Ins}-\bar X_{No.Ins}}{\sqrt{\frac{s^2_{Ins}}{n_{Ins}}+\frac{s^2_{No. Ins}}{n_{No. Ins}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

$df = n_{ins} +n_{No Ins}-2=16+16-2 =30$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{30}>2.385)=0.0236$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

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I think that the missing word here is:
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