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7. A sample of 106 body temperatures was found to have a mean of [tex]$98.20^{\circ} F$[/tex]. Assume that [tex]$\sigma$[/tex] is known to be [tex]$0.62^{\circ} F$[/tex] and a confidence level of [tex]$a=0.05$[/tex] is used.

Find the value of the test statistic [tex]$z$[/tex] that you would use to test the claim that the mean body temperature of the population is equal to [tex]$98.6^{\circ} F$[/tex] using:

[tex]\[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\frac{\sigma}{\sqrt{n}}} \][/tex]

A. [tex]$-6.64$[/tex]
B. 6.64
C. [tex]$-5.23$[/tex]
D. 0.08


Sagot :

To find the value of the test statistic [tex]\( z \)[/tex] for the hypothesis test, we'll follow these steps:

1. Identify the given values:
- Sample size ([tex]\( n \)[/tex]) = 106
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 98.20°F
- Population mean ([tex]\( \mu \)[/tex]) = 98.6°F
- Population standard deviation ([tex]\( \sigma \)[/tex]) = 0.62°F

2. Write down the formula for the test statistic [tex]\( z \)[/tex]:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

3. Plug the given values into the formula:

- The numerator [tex]\( \bar{x} - \mu \)[/tex] is:
[tex]\[ 98.20 - 98.6 = -0.4 \][/tex]

- The denominator [tex]\( \frac{\sigma}{\sqrt{n}} \)[/tex] is:
[tex]\[ \frac{0.62}{\sqrt{106}} \][/tex]

4. Compute the denominator:
[tex]\[ \frac{0.62}{\sqrt{106}} \approx \frac{0.62}{10.2956} \approx 0.0602 \][/tex]

5. Compute the [tex]\( z \)[/tex]-value:
[tex]\[ z = \frac{-0.4}{0.0602} \approx -6.64 \][/tex]

So, the value of the test statistic [tex]\( z \)[/tex] is approximately -6.64.

Among the given options:
- [tex]\( -6.64 \)[/tex]
- [tex]\( 6.64 \)[/tex]
- [tex]\( -5.23 \)[/tex]
- [tex]\( 0.08 \)[/tex]

The correct value for the test statistic [tex]\( z \)[/tex] is:
[tex]\[ -6.64 \][/tex]