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Notes: On Question 1, assume we are conducting the z-test with a p-level of 0.05

ID: 3204672 • Letter: N

Question

Notes:

On Question 1, assume we are conducting the z-test with a p-level of 0.05. Thus, you need to get a z-value of 1.96 to say that we have a statistically significant result. Typically, if no p-value is given, you should assume the critical p-value is 0.05.

Reminder: the z-value of 1.96 corresponds to a percentage of 47.5% on Table A in Tanner. This means that between the center of the distribution and the z-value of 1.96, 47.5% of the distribution is covered. So, 2.5% of the distribution is left in the tail. If our p-value is set at 0.05, this means we will only reject the null hypothesis when we think our sample fell in the extreme 5% of the population distribution. Thus, if our population falls in the upper 2.5% or the lower 2.5%, then we reject null and say we have a statistically significant result.

On Question 2, you need to calculate a 95% confidence interval and a 99% confidence interval. For both calculations, you will need to have a z-value (this is a necessary part of the formula). For 95%, the z-value is 1.96. For the 99% confidence interval, the z-value is 2.58. Again, these are both drawn from Table A. We discussed this in class, but I think we may have gone over it rather fast. We will discuss again this week. But, you should be on the right track if the range of values covered by your 99% confidence interval is WIDER than the range covered by the 95% interval.

EDUC 15: Problem Set 3 This problem set is due in class on Tuesday, February 7. Legibly write your name (or your group member's names at the top of each page. Fill in your answers in the spaces provided. If you need more room, attach and clearly label an extra sheet to this document When appropriate, show your work! 1. (10 points) The personnel director of UCI school of Education determines the keyboarding speeds of a random sample of secretaries from her school. She wishes to test the hypothesis that the mean for her population is equal to 50 words per minute (ku 50), the national mean for secretaries. The personnel director obtained a mean M 48 words per minute based on a sample of size 36. Further, she knows that the standard deviation of keyboarding speeds is 10 (o 100. She wishes to use the criterion of a level of significance p-0.05 A. State the null hypothesis and alternative hypothesis for this problem (both in symbols and in words, no more than 2 sentences) B. Explain in your own words what a level of significance of p 0.05 means. Think about the underlying distributions implied by Ho and Ha. LIMIT YOUR RESPONSE TO NO MORE THAN 3 SENTENCES. C. Calculate the standard error of the mean (om). SHOW YOUR WORK. D. Calculate the value of the z test. SHOW YOUR WORK. E. Explain the decision or conclusion that the personnel director would make. LIMIT YOUR RESPONSE TO NO MORE THAN TWO SENTENCES.

Explanation / Answer

Q3.
A.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=33.1
Standard deviation( sd )=4
Sample Size(n)=36
Confidence Interval = [ 33.1 ± Z a/2 ( 4/ Sqrt ( 36) ) ]
= [ 33.1 - 1.96 * (0.67) , 33.1 + 1.96 * (0.67) ]
= [ 31.79,34.41 ]
B.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=33.1
Standard deviation( sd )=4
Sample Size(n)=36
Confidence Interval = [ 33.1 ± Z a/2 ( 4/ Sqrt ( 36) ) ]
= [ 33.1 - 2.58 * (0.67) , 33.1 + 2.58 * (0.67) ]
= [ 31.38,34.82 ]
C.
Increaseing the confidence interval widens the interval
D.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=33.1
Standard deviation( sd )=4
Sample Size(n)=100
Confidence Interval = [ 33.1 ± Z a/2 ( 4/ Sqrt ( 100) ) ]
= [ 33.1 - 1.96 * (0.4) , 33.1 + 1.96 * (0.4) ]
= [ 32.32,33.88 ]
E.
Increasing the sample size narrowing the interval

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