Question

3. Presently, the mean life expectancy of a rare strain of bacteria is 12 hours. A scientist claims that she has developed a medium and therefore disagrees with this hypothesis that it is not 12 hours. The scientist tests 25 cultures and finds that they have a mean life of 13.25 hours with a sample standard eviation = 2 hour. Do these results show that the medium is effective in increasing the bacteria's life expectancy? (Use <= 5%) a. State the null and alternative hypothesis b. Determine and draw the hypothesis test model (Label everything) c. Determine the critical value(s) and test statistic d. Determine the conclusion 4. The probability that Casey, a softball player gets a hit during a single time at bat is 3/10. If she goes to bat 6 times in her next game, determine the probability to four decimal places a. That she gets at least 3 hits in the game. b. That she gets at most 4 hits in the game c. That she does not get a hit in the game. d. That she gets exactly 3 hits in the game

Answer 1

3.

There is sufficient evidence to support the claim that the medium is effective in increasing the bacteria's life expectancy.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho pe=12 against alternative hypothesis H : > 12 Let e denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, = 13.250, 82 = 4.000000, s = 2.00, n=25 The test statistic can be written as: /(– 12) te s which under H, follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05,n-1 = 1.710882 0.4 0.3 0.2 A 0.1 - 0.0 - -2 0 t Now, Value of the test statistic : The value of the test statistic tobs = V25 (13.2500 - 12) 2.0000 = 3.125000 ~ 3.12 associated degrees of freedom = 24 The critical value = 1.710882 1.71 P-value = P(t24 > 3.125000) =1- P(t24 <3.125000) = 1-0.997699 = 0.002301 0.0023 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.710882, so we reject H, at 0.05 level of significance Hence, we can conclude that population mean is significantly greater than 12

4.

Let X denotes the number of times that she hits among 6 times.

Here,

X ~ Binomial(n = 6, p = 3/10 = 0.3)

The probability mass function of X is

P(X = z) = *0.3" * (1 – 0.3)6-1, x = 0,1,2,3,4,5,6

a) $P(X \geq 3) =1 - P(X<3) = 1 - P(X\leq 2) = 1 -\sum_{x=0}^{2} \binom{6}{x}*0.3^x*(1-0.3)^{6-x} = 1 - 0.74431 =0.25569$ b)

$P(X \leq 4) =\sum_{x=0}^{4} \binom{6}{x}*0.3^x*(1-0.3)^{6-x} = 0.989065 \approx 0.9891$

c)

$P(X =0) =\binom{6}{0}*0.3^0*(1-0.3)^{6-0} = 0.117649 \approx 0.1176$

d)

$P(X =3) =\binom{6}{3}*0.3^{3}*(1-0.3)^{6-3} = 0.18522$