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2 years ago

The heights of a group of boys and girls at a local middle school are shown on the dot plots below.

Mathematics

2 answers:

sammy [17]2 years ago

8 0

Answer:

D) The girls are generally taller than the boys.

Step-by-step explanation:

The heights of a group of boys and girls at a local middle school are shown on the dot plots. When comparing the shapes of the two sets of data, the girls are generally taller than the boys.

netineya [11]2 years ago

4 0

<u>Answer:</u>

The most appropriate answer option is D) The girls are generally taller than the boys.

<u>Step-by-step explanation:</u>

A) The shortest boy is taller than the shortest girl:

The shortest guy is not taller than the girl so its false.

B) The range for the girls is greater than the range for the boys:

Range for girls = 56 - 44 = 12

Range for boys = 54 - 41 = 13

C) There is an outlier in the data for the boys, but not for the girls:

It is present in both.

D) The girls are generally taller than the boys:

Average height of boys = $\frac{(41)+(44\times3)+(46\times3)+(48\times2)+(50\times3)+(52\times)+(54\times4)}{20}$ = 49.05

Average height of girls = $\frac{(44\times2)+(46\times2)+(48)+(50\times3)+(52\times4)+(54\times3)+(56\times4)}{20}$ = 50.9

Therefore, the correct answer option is D) The girls are generally taller than the boys.

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2 years ago

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d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

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

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

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

$p_v =P(t_{14}>1.516)=0.076$

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3 years ago