Question

A Nadimul Haque - σ Stat 20 1 Test #2 Chap 9-11.doc - Read-Only-Compatibility Mode Insert Draw Design LayoutReferences Mailings Review View Help lell me what you want to do File Home | Share AabCc AaBbC AaBbCcl AaBbCel AaBbCcD AaBbC.Relace Arial , x, A뽀 . . .ミ : . g«b_ Emphasis Heading i 11 Normal! Strong Share Upload Paste Copy Title : Subtitle Format Painter B 1 u . abex Select. Editing Font Styles Bax The US Census Bureau reports that the average number of employees in a small business is 16.1. Suppose a sample of 49 small businesses showed a mean of 15 employees, with a standard deviation of 25. If appropriate, test whether the population mean number af employees in a small business is different fram 16.1 using a level of significance α-001 4.2 The National Center for Health Statistics reported: "Childbearing by unmarried women increased to record levels for the nation in 2005.In that year, 36.8% of all births were to unmarried women. Suppose that a random sample taken this year of 1000 births showed 380 to unmarried women. If appropriate, test whether the population proportion has increased since 2005, using level of significance α- 4.3 A reduction in heart rate variability is associated with elevated levels of stress, since the body continues to purmp adrenaline after high-stress situations, even when at rest. Suppose the standard deviation of heartbeats in the general population is 20 beats per minute, and that the population of heart rates is normal. A random sample of 50 individuals leading high-stress lives has a sample variance of 200 beats per minutes. If appropriate, test using level of significance a 0.05 whether the population standard deviation for those leading high-stress lives is lower than that in the general population. EE 匡晒ー Page 4 18 1210 words L 5:27 PM 5/5/2019

Answer 1

4.1)

Sample size = n = 49

Sample mean = $\bar{x}$ = 15

Standard deviation = s = 25

Claim: The population mean number of employees in a small business is different from 16.1

The null and alternative hypothesis is

$H0:\mu=16.1$

$H1:\mu\neq 16.1$

Level of significance = 0.01

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

Vn

$t=\frac{15-16.1 }{\frac{25}{\sqrt{49}}}=\frac{-1.1}{3.571429}=-0.31$

Degrees of freedom = n - 1 = 49 - 1 = 48

Critical value = 2.682    ( Using t table)

Test statistic | t | < critical value we fail to reject null hypothesis.

Conclusion:

The population mean number of employees in a small business is NOT different from 16.1

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