A variation of the Student's t-test used to determine if two sample means are substantially different is known as the two sample T-test (also known as Welch's t-test, Welch's adjusted T, or unequal variances t-test). The test's degrees of freedom have been changed, which generally boosts the test's power for samples with unequal variance.

T-Test Statistic for The Procedure

It is given that mean of the differences = 1.33

And, the standard deviation of the differences, standard error = 2.90

The test static formula is given as,

t = (estimate - hypothized value) / standard error

Here, t is the t- test statistic, and estimate is the mean of the differences.

Since it is given that all conditions are met, the hypothized value can be taken as 0

Substituting substituting estimate = 1.33 and standard error = 2.90 in the t-test statistic formula, we get,

t = (1.33-0)/2.90

or t = 0.458

learn more about t-test here:

brainly.com/question/14128303

#SPJ4

8 0

1 year ago

Help me with 24,26,28, and 30 please.

Novay_Z [31]

24)
The number 1/10 can be expressed as a decimal by dividing the top number by the bottom.
1/10=0.1
0.8>0.1
0.8>1/10

26)
The number 3/5 can also be expressed as a decimal, using the method I described for 24.
3/5=0.6
0.6>0.35
3/5>0.35

28)
The number 2/5 can also be expressed as a decimal.
2/5=0.4
0.4>0.3
2/5>0.3

30)
The number 1/6 cannot be expressed as a decimal. So, express 0.2 as a fraction.
0.2=2/10=1/5
1/5=6/30
1/6=5/30
6/30>5/30
0.2>1/6

Hope this helps!

8 0

3 years ago

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st

Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .

What is the probability that a line width is greater than 0.62 micrometer?

That is $P(X > 0.62)$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.62 - 0.5}{0.05}$

$Z = 2.4$

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

$P(X \leq 0.62) + P(X > 0.62) = 1$

$P(X > 0.62) = 1 - 0.99180 = 0.0082$

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

3 0

2 years ago

Broderick had $420 in his checking account. He made 6 deposits of $35.50 each. He needs to write 4 checks for $310.75 each.How m

kvasek [131]

So, first you would need to multiply 6 by $35.50, then take that number and add it to the $420 to get $633.00. Then you need to multiply $310.75 by 4. After take that $1243.00 and subtract $633.00 from it to get $610.00. So, the answer is: Broderick would need to add at least $610 to his account to cover his checks.

3 0

3 years ago