A teacher wants to see if a new unit on taking square roots is helping students learn. She has five randomly selected students t
2 answers:
Answer:
The P-value for this analysis is P=0.00006.
Step-by-step explanation:
We are conducting a matched-pair test, with a calculated test statistic of t=-14.9.
The degree of freedom for a sample size of 5 students is:
For t=-14.9 with 4 degrees of freedom, a left-tail test yields a P-value of:
This test aims to verify the claim that the new unit regarding square roots aids student learning. The data suggests statistical support that the new curriculum improves learning outcomes for students.
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Answer:
When x = 72, the z score is:
The average is 58
This z-score indicates that x= 72 is 1.273 standard deviations above the mean.
Step-by-step explanation:
Let’s consider the following details for this question: John's typing speed on a test is assumed to follow a normal distribution. Let X denote the number of words he can type in a minute. Hence, X ~ N(58, 11). It’s important to round the result to three decimal points if needed.
Provide your response below for words per minute on a typing test conducted on Sunday. The z score when x = 72 is
In this scenario, we acknowledge that the variable under consideration is represented by a normal distribution:
The formula for the z score is as follows:
By substituting values, we derive:
The mean is 58
This z-score indicates that x= 72 is 1.273 standard deviations higher than the mean.
The solution to the query
