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Chapter 15

# Chapter 15.odt

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Simon Fraser University

Statistics

STAT 101

Qian( Michelle) Zhou

Fall

Description

Chapter 15 Test of Significance: The Basics
Stating Hypothesis
• null hypothesis (H0) – the claim tested by a statistical test; designed to assess strength of evidence against null
hypothesis. Is usually a statement of “no effect” or “no difference”
◦ H0: μ = 0
• alternative hypothesis (Ha) – claim about population; trying to find evidence FOR Ha
◦ one sided if Ha > or < 0 (the null hypothesis value
▪ Ha: μ >0 or Ha: μ<0
◦ two sided if parameters is different from the null value Ha: μ ≠ 0
▪ it just means that as long as it's not 0 (or any number that is H0) then Ha can be on either side
P-Value and Statistical Significance
• probability that measure the strength evidence against a null hypothesis is called a P-Value
◦ smaller the p-value, the stronger the evidence against H0 is provided by data
• Test statistic – calculated from sample data measures how far the data diverge from what we would expect if
the null hypothesis H0 were true. Large values of statistics show that the data are not consistent w/ H0
◦ Z = (x̄ - μ0) / (σ/√n) ~ N(0,1) if H0 is ture
• if we fail to find evidence against H0, it doesn't mean H0 is true, it's just that that data are not inconsistent with
H0
• how small a p-value we should require to reject H0
◦ significance level (α)
◦ if Pvalue < α , we can reject H0
▪ ex) if P-value < 0.05, say “ that data are significant”
▪ ex) if Pvalue < 0.01 say, the “data are highly significant” etc
◦ if Pvalue is ≥ α , we fail to reject H0
Test for a Population Mean
• data that would rarely occur if the null hypothesis H0 were true provide evidence that H0 is not true. P-value
gives us the probability that it would rarely occur
Example 1) Abag of chips says that theres 200 chips in the bag. But we don't think that's true. We hypothesize that
theres less than 200. We examined 5 bags of chips, w/ standard deviation of 15.65. We also find that the average
number of chips in all 5 bags 181. is H0 true? With significance level of .-5.
H0: μ = 200 Ha: μ < 200
Z = (x̄ - μ0) / (σ/√n) = 181 – 200/ 15.65/sqrt5 = -2.71 --> table a --> P value = 0.0034
Since Pvalue < a , we can reject H0, that 200 is the average amount of chips in the bag. This is true if Ha < μ. If we
want to find Ha > μ....it'll be a different calculation.Assume values are same for all values from above.. but Ha > μ.
We find Pvalue is 0.0034, but since we're looking for area RIGHT of μ, it is 1-p value
1-0.0034 = 0.9966. Comparing this pvalue to significance level. This is bigger than a. So we fail to reject H0.
Example 2) Astudy says that university students study 9.5 hours a week. We, for some reason, don't think it's true, but
we don't know if students study more than or less than 9.5 hours The SD is 2.94 hours and we sampled 5 students. We
have an average of 7 hours with significance level of 0.05.
H0: μ = 9.5 Ha: μ ≠ 9.5
Z = (x̄ - μ0) / (σ/√n)
= 7-9.5/ 2.94/sqrt5 = -1.90 --> TableA --> Pvalue = 0.0287. Since it is Ha: μ ≠ 9.5...pvalue is 2(Pvalue = .0287)
2 Pvalue = .0574. Since Pvalue > a, we fail to reject H0
What if we want to find it using Confidence interval??
Remember that x̄ ± z*(σ/√n)
z* (look at where the 2sided P is, and then look up at the z*, that's your z*) = 1.960
7± 1.960(1.314)
Range from 4.423 – 9.557.And Since 9.5 hours falls in the range of 4.423 – 9.557, we fail to reject H0.
Example 15.6 Executives blood pressure
– reports of blood pressure for males 35-44 years old, has mean 128 and SD 15. Looks at 72 executives. x̄ is
126.07. Is this evidence that the comp

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