Z test for two sample mean
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This document explains how to perform a z-test for two sample means. The z-test can be used to determine if the difference between the means of two random samples from a normal population is statistically significant. The document provides the formula for the z-test and walks through an example comparing the average monthly family incomes from two neighborhoods based on random samples. It demonstrates stating the null and alternative hypotheses, performing the z-test calculation, comparing the results to the critical value, and making a conclusion about rejecting or failing to reject the null hypothesis based on the level of significance.
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Z test for two sample mean
- 3. Z-TEST FOR TWO SAMPLE MEAN • Is used when comparing two separate samples drawn at random taken from a normal population • Used to test whether the difference between the two values of 𝒙 𝟏 and 𝒙 𝟐 is significant or can be attributed to chance.
- 4. Formula: Where: = the mean of sample 1 = the mean of sample 2 = the standard deviation of sample 1 = the standard deviation of sample 2 = the size of sample 1 = the size of sample 2 𝑿 𝟏 𝑿 𝟐 𝑺 𝟏 𝑺 𝟐 𝒏 𝟏 𝒏 𝟐
- 5. EXAMPLE • A bank is opening a new branch in one of two neighborhoods. One of the factors considered by the bank is whether the average monthly family (in thousand pesos) in the two neighborhoods differed. From census records, the bank drew two random samples of 100 families each and obtained the following information. SAMPLE A SAMPLE B 𝑿 𝟏 = 𝟏𝟎, 𝟖𝟎𝟎 𝑿 𝟐 = 𝟏𝟎, 𝟑𝟎𝟎 𝑺 𝟏 = 𝟑𝟎𝟎 𝑺 𝟐 = 𝟒𝟎𝟎 𝒏 𝟏 = 𝟏𝟎𝟎 𝒏 𝟐 = 𝟏𝟎𝟎
- 6. SOLUTION (USING ONE TAILED TEST) • I. State the null and alternative hypothesis. • A. 𝑯 𝒐 = The average monthly family income in the two neighborhoods, A and B are equal. • 𝑯 𝒐: 𝝁 𝟏 = 𝝁 𝟐 • B. 𝑯 𝒂 = The average monthly family income of neighborhood A is higher than neighborhood B. • 𝑯 𝒂: 𝝁 𝟏 > 𝝁 𝟐
- 7. SOLUTION • II. Set the level of significance. • 𝜶 = 𝟎. 𝟎𝟓 III. Test Statistic Z-test for Two Sample Mean
- 8. SOLUTION • IV. Computation • Z= 𝑿 𝟏− 𝑿 𝟐 (𝑺 𝟏) 𝟐 𝒏 𝟏 + (𝑺 𝟐) 𝟐 𝒏 𝟐 = 𝟏𝟎,𝟖𝟎𝟎−𝟏𝟎,𝟑𝟎𝟎 (𝟑𝟎𝟎) 𝟐 𝟏𝟎𝟎 + (𝟒𝟎𝟎) 𝟐 𝟏𝟎𝟎 = 𝟓𝟎𝟎 𝟗𝟎𝟎+𝟏𝟔𝟎𝟎 = 𝟓𝟎𝟎 𝟓𝟎 = 𝟏𝟎 SAMPLE A SAMPLE B 𝑿 𝟏 = 𝟏𝟎, 𝟖𝟎𝟎 𝑿 𝟐 = 𝟏𝟎, 𝟑𝟎𝟎 𝑺 𝟏 = 𝟑𝟎𝟎 𝑺 𝟐 = 𝟒𝟎𝟎 𝒏 𝟏 = 𝟏𝟎𝟎 𝒏 𝟐 = 𝟏𝟎𝟎
- 9. SOLUTION • V. Critical/Tabular Value • 𝒛 𝒕𝒂𝒃 = ±𝟏. 𝟔𝟓 (𝜶 = 𝟎. 𝟎𝟓)one tailed • VI. Decision Rule • Since the 𝒛 𝒄𝒐𝒎𝒑𝒖𝒕𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 is greater than the 𝒛 𝒕𝒂𝒃𝒖𝒍𝒂𝒓 𝒗𝒂𝒍𝒖𝒆, reject the null hypothesis and accept the alternative hypothesis.
- 10. SOLUTION • VII. Conclusion • Since the 𝒛 𝒄𝒐𝒎𝒑𝒖𝒕𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 of 10 is greater than the 𝒛 𝒕𝒂𝒃𝒖𝒍𝒂𝒓 𝒗𝒂𝒍𝒖𝒆 of 1.65, reject the null hypothesis and accept the alternative hypothesis, regardless of sign at 0.05 level of significance. The research hypothesis confirms that the average family monthly income of neighborhood A is higher than neighborhood B.
- 11. ACTIVITY! TRY THIS • 1. The Brand A cellphone company claims that its phones have an almost the same lifespan as of the Brand B cellphone company. A test is conducted to validate whether the claim is true. 40 phones from Brand A and 70 phones from Brand B are chosen for testing, and were found that the mean life for each is 32 months for Brand A and 35 months for Brand B. The standard deviation for Brand A is 5 months while 7 months for Brand B. Test the hypothesis if the average life of an android phone from Brand A is less than the average life of android phone from Brand B using a level of significance of 0.01