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nikdorinn [45]
3 years ago
10

Which statement is true about the variability of the distributions? The interquartile range of the typing speeds after the cours

e is greater than the interquartile range of the speeds before the course. The interquartile ranges of the two distributions are the same. The range of the speeds after the course is smaller than the range of the speeds before the course. The ranges of the two distributions are the same.
Mathematics
2 answers:
nevsk [136]3 years ago
7 0

Answer:The range of the speeds after the course is smaller than the range of the speeds before the course.

Step-by-step explanation:

DochEvi [55]3 years ago
6 0

Answer: Your answer is: C. The range of the speeds after the course is smaller than the range of the speeds before the course.

I hope it will help you!!

Your Welcome

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Given the following information about a hypothesis test of the difference between two means based on independent random samples,
HACTEHA [7]

Answer:

S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143

And the deviation would be just the square root of the variance:

S_p=4.259

Then the statistic is given by:

t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674

And the correct option would be:

t = 1.674

Step-by-step explanation:

Data given:

n_1 =13 represent the sample size for group 1

n_2 =10 represent the sample size for group 2

\bar X_1 =12 represent the sample mean for the group 1

\bar X_2 =9 represent the sample mean for the group 2

s_1=5 represent the sample standard deviation for group 1

s_2=3 represent the sample standard deviation for group 2

We are assuming two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

The pooled variance is given by:

S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143

And the deviation would be just the square root of the variance:

S_p=4.259

Then the statistic is given by:

t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674

And the correct option would be:

t = 1.674

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4 years ago
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