Hypothesis about the Difference of two Population Means Lecture 32

 Hypothesis about the Difference of two Population Means

Lecture 32

To test the hypothesis about the difference between population means. The following cases are to be considered.

Case 1: Hypothesis about the Difference between Population Means when σ1, σ2 are known.

To test the null hypothesis that the difference between the two population means is zero or some specified value. Let two independent random samples of size n1, n2 selected from two normal populations having means μ1, μ2 and known standard deviations σ1, σ2. Let X¯1, X¯2 be the estimates of μ1, μ2 computed from the values of the selected samples. The sampling distribution of (X¯1 - X¯2) approaches normal distribution. 

Suppose it is desired to test ${\mu }_{\mathrm{1\; -}} {\mu }_2 =$△∘

Testing Procedure:

i. State the null and alternative hypothesis

H0: ${\mu }_{\mathrm{1\; -}} {\mu }_2 =$△∘ Vs. H1: ${\mu }_{\mathrm{1\; -}} {\mu }_2 \ne$△∘

H0: ${\mu }_{\mathrm{1\; -}} {\mu }_2$≥ △∘ Vs. H1: ${\mu }_{\mathrm{1\; -}} {\mu }_2 <$△∘

H0: ${\mu }_{\mathrm{1\; -}} {\mu }_2$≤ △∘ Vs. H1: ${\mu }_{\mathrm{1\; -}} {\mu }_2 >$△∘

ii. The Significance Level; α 

iii. The test statistic: As  σ1, σ2 is known.

under H0.

iv. Critical Region:

Reject H0 when |z| ≥ zα/2

Reject H0 when z < - zα

Reject H0 when z > zα

v. Computation:

vi. Remarks

Case 2: Hypothesis about the Difference between Population Means when σ1, σ2 are unknown.

To test the null hypothesis that the difference between the two population means is zero or some specified value. Let two independent random samples of size n1, n2 selected from two normal populations having means μ1, μ2 and unknown standard deviations σ1, σ2. As σ1, σ2 are unknown, so replace by its estimates S1, S2. Let X¯1, X¯2 be the estimates of μ1, μ2 computed from the values of the selected samples. If the sample sizes are large, then the sampling distribution of (X¯1 - X¯2) approaches normal distribution. 

Suppose it is desired to test ${\mu }_{\mathrm{1\; -}} {\mu }_2 =$△∘

The testing procedure remains the same.

Example 8.13: Ten percent of the BS botany & BS zoology students are chosen in order to check whether the two departments’ performance in the statistics course is the same or different. Two independent random samples of 12 and 15 students each are chosen from the BS botany and BS zoology programs. Such kind of research was also conducted in the past in both of the departments, with the standard deviations of BS botany and BS zoology being 1.5 and 2.13, respectively. Two random independent samples of size 12 and 15 students are selected from BS botany & BS zoology, respectively. The sample average GPAs of BS botany & zoology are 3.40 and 3.67, respectively. Test the claim that the performance of students in the statistics course of both departments is identical at the 5% significance level.

Solution: Let μ1 & μ2 represent the average performance of students of the BS botany & BS zoology departments. We setup our null & alternative hypotheses as:

i. State the null and alternative hypothesis

H0: ${\mu }_{\mathrm{1\; -}} {\mu }_2 =\; 0$ Vs. H1: ${\mu }_{\mathrm{1\; -}} {\mu }_2 \ne \; 0$

$<span\; style="font-family:\; "Times\; New\; Roman",\; serif;\; font-size:\; 12pt;"><b>Example\; 8.14</b>:\; A\; </span><a\; name="\_Hlk188174342"\; style="font-family:\; "Times\; New\; Roman",\; serif;\; font-size:\; 12pt;">market\; investigator\; claims\; </a><span\; style="font-family:\; "Times\; New\; Roman",\; serif;\; font-size:\; 12pt;">that\; the\; average\; sale\; of\; a\; corner\; shop\; is\; more\; than\; the\; non-corner\; shop.\; To\; check\; the\; </span><a\; name="\_Hlk188174702"\; style="font-family:\; "Times\; New\; Roman",\; serif;\; font-size:\; 12pt;">market\; investigator</a><span\; style="font-family:\; "Times\; New\; Roman",\; serif;\; font-size:\; 12pt;">,\; claim\; two\; random\; and\; independent\; samples\; each\; of\; size\; 60\; days\; of\; the\; corner\; and\; non-corner\; shops\; gave\; the\; average\; sale\; is\; \$100\; and\; \$85\; with\; a\; standard\; deviation\; of\; \$5.40\; and\; \$6.23.\; The\; claim\; of\; the\; market\; investigator\; is\; validated\; at\; the\; 0.5\%\; significance\; level.</span>$

Solution: Let μ1 & μ2 represent the average sales of corner and non-corner shops. We setup our null & alternative hypotheses as:

$$i. State the null and alternative hypothesis

H0: ${\mu }_{\mathrm{1 }}$≤${\mu }_2$ Vs. H1: ${\mu }_{\mathrm{1\; >}} {\mu }_2$

${\mathrm{or\; we\; can\; state\; as: }}_{}$

i. State the null and alternative hypothesis

H0: ${\mu }_{\mathrm{1\; -}} {\mu }_2 =\; 0$ Vs. H1: ${\mu }_{\mathrm{1\; -}} {\mu }_2 \ne \; 0$

Solution: Let P1 and P2 be the proportion of defective LED TVs produced by companies A and B, respectively.

i. H0: P1 - P2 = 0 Vs. H1: P1 - P2 ≠ 0

ii. The significance level; α = 0.05

iii. The test statistic: As n1 & n2 are large enough, then

iv. Critical Region:

Reject H0 if |z| ≥ z0.05/2 = 1.96

v. Computation:

vi. Remarks: The computed z values fall in the acceptance region; the sample data does not provide sufficient evidence to reject the null hypothesis at the 5% significance level. Thus, it is concluded that the proportion of defective LED TVs produced by both companies is the same.

Example 8.17: The purpose of the study is to compare the proportions of adult patient responses to two different types of flue treatment. Twenty of the 200 people assigned to Medicine A respond in better condition. Then medicine B is prescribed; the remaining 200 patients exhibit the same flu-like symptoms, while 12 respond better. According to a doctor, drug A works 10% better than drug B.

Test at a 1% level of significance.

Solution:

i. H0: P1 - P2 = 0.10 Vs. H1: P1 - P2 ≠ 0.10

ii. The significance level; α = 0.01

iii. The test statistic: As n1 & n2 are large enough, then

iv. Critical Region:

Reject H0 if |z| ≥ z0.01/2 = 2.5758

v. Computation:

vi. Remarks: The computed z value falls in the acceptance region; the sample data does not provide sufficient evidence to reject the null hypothesis. It is concluded that the drug A is 10% more effective than drug B.

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