Question

A low-noise transistor for use in computing products is being developed. It is claimed that the mean noise level will be below the 3.00 dB level of products currently in use. (Assume that noise level is normally distributed).
i. Set up the appropriate null and alternative hypotheses for verifying the claim.
ii. Find the critical point for 99% confidence test based on a sample of size 20. Using the values 2.2 and 0.88 of sample mean and sample standard deviation, respectively, to test null hypothesis against alternative hypothesis defined in (i) at  = 0.01. Justify.

Answer 1

Answer on Question #46541 – Math -Statistics and Probability

A low-noise transistor for use in computing products is being developed. It is claimed that the mean noise level will be below the $\mu_0 = 3.00\ dB$ level of products currently in use. (Assume that noise level is normally distributed).

i. Set up the appropriate null and alternative hypotheses for verifying the claim.

ii. Find the critical point for $99\%$ confidence test based on a sample of size $n = 20$. Using the values $\bar{x} = 2.2$ and $s = 0.88$ of sample mean and sample standard deviation, respectively, to test null hypothesis against alternative hypothesis defined in (i) at $\alpha = 0.01$. Justify.

Solution

i. $H_0: \mu \leq \mu_0 = 3.00; H_\alpha: \mu > \mu_0$.

ii. The significance level $\alpha = 0.01$.

It is an one-sided test. Number of degrees of freedom $df = 20 - 1 = 19$. $t_{crit} = t_{0.01;19} = 2.86$.

$$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{2.2 - 3}{\frac{0.88}{\sqrt{20}}} = -4.065.$$

Don't reject $H_0$ because $t = -4.065 < t_{crit} = 2.86$.

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