A two sample (independent-sample) study with n = 6 in each sample, produces a sample mean difference of 4 points and a pooled variance of 12. What is the value for the sample t score?Group of answer choices

A)1

B)2

C)4/6

D)4/8

A)1

B)2

C)4/6

D)4/8

smuklica8i
2022-07-22
Answered

A two sample (independent-sample) study with n = 6 in each sample, produces a sample mean difference of 4 points and a pooled variance of 12. What is the value for the sample t score?Group of answer choices

A)1

B)2

C)4/6

D)4/8

A)1

B)2

C)4/6

D)4/8

You can still ask an expert for help

Jazlene Dickson

Answered 2022-07-23
Author has **15** answers

The pooled variancce = 12

The pooled standard deviation = $\sqrt{12}$

The formula for Standard error using pooled standard deviation is

$SE={S}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{12}\sqrt{\frac{1}{6}+\frac{1}{6}}\phantom{\rule{0ex}{0ex}}=\sqrt{12}\times \sqrt{\frac{1}{3}}\phantom{\rule{0ex}{0ex}}=\sqrt{4}=2$

The formula for sample t-score is given by

$t=\frac{\overline{x}-\mu}{SE}$

where $\overline{x}=$ sample mean

$\mu =mean$

$t=\frac{4}{2}\phantom{\rule{0ex}{0ex}}=2$

Option B is the right choice

The pooled standard deviation = $\sqrt{12}$

The formula for Standard error using pooled standard deviation is

$SE={S}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{12}\sqrt{\frac{1}{6}+\frac{1}{6}}\phantom{\rule{0ex}{0ex}}=\sqrt{12}\times \sqrt{\frac{1}{3}}\phantom{\rule{0ex}{0ex}}=\sqrt{4}=2$

The formula for sample t-score is given by

$t=\frac{\overline{x}-\mu}{SE}$

where $\overline{x}=$ sample mean

$\mu =mean$

$t=\frac{4}{2}\phantom{\rule{0ex}{0ex}}=2$

Option B is the right choice

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I am struggling with the following question:

A test is constructed to see if a coin is biased. It is tossed 10 times and if there are 10 heads, 9 heads, 1 head or 0 heads, it is declared to be biased. Can 20 be the significance level of this test?

My thinking is as follows: ${H}_{0}:p=0.5$

${H}_{1}:p\ne 0.5$

Let X be the number of heads, under ${H}_{0}$, X$\sim $B(10,0.5).

If we look at a table of values, we get:

X=0,P=0.00098

X=1,P=0.0107

X=9,P=0.999

X=10,P=1

Since it is declared biased for these values, P<0.1 or P>0.9. Therefore, shouldn’t we be able to use 20% as the significance level, yet my book says we can’t. Any clarification?

A test is constructed to see if a coin is biased. It is tossed 10 times and if there are 10 heads, 9 heads, 1 head or 0 heads, it is declared to be biased. Can 20 be the significance level of this test?

My thinking is as follows: ${H}_{0}:p=0.5$

${H}_{1}:p\ne 0.5$

Let X be the number of heads, under ${H}_{0}$, X$\sim $B(10,0.5).

If we look at a table of values, we get:

X=0,P=0.00098

X=1,P=0.0107

X=9,P=0.999

X=10,P=1

Since it is declared biased for these values, P<0.1 or P>0.9. Therefore, shouldn’t we be able to use 20% as the significance level, yet my book says we can’t. Any clarification?

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We want to make an hypothesis test for the mean value $\mu $ of a normal population with known variance ${\sigma}^{2}=13456$, using a sample of size $n=100$ that has sample mean value equal to $562$.

Calculate the $p$-value.

Make the test with significance level $1\mathrm{\%}$ about if the population mean value from which the sample comes from is greater than $530$ using the $p$-value.

For the first one, about the $p$-value, do we have to calculate $P\left(\frac{530-562}{\frac{\sigma}{\sqrt{n}}}\right)$?

And for the second we have to check the $p$-value with the significance level, right?

Calculate the $p$-value.

Make the test with significance level $1\mathrm{\%}$ about if the population mean value from which the sample comes from is greater than $530$ using the $p$-value.

For the first one, about the $p$-value, do we have to calculate $P\left(\frac{530-562}{\frac{\sigma}{\sqrt{n}}}\right)$?

And for the second we have to check the $p$-value with the significance level, right?