Confidence Interval of the Mean......?

Question

Below are the weights (g) of two-month old catfish caught in a pond

157 165 200 167 200 185 192 173 174 176

203 209 185 163 190 170 174 164 201 181

185 169 183 156 168 172 180 170 175 182

178 193 183 159 194 191 165 165 196 192

150 149 162 189 195 172 166 202 171 190

Estimate the 95% and 99% confidence interval of the mean of the weights.

For the current year, the catfish were provided with a new feed formula. the mean weight of two-month old fish was found to be 195g. Based on these information, what can you comment on the effects of the new feed on the growth of the catfish?

Please help me solve it...........

Answer 1

ANSWER: Sample of 50 two-month old catfish with new feed formula shows to a 99% confidence the "true mean" weight is affected by new feed formula on two-month old catfish.

Why???

SINGLE SAMPLE TEST, TWO-TAILED, 7 - Step Procedure for t Distributions

1. Parameter of interest: "μ" = population mean weights of two-month old catfish.

2. Null hypothesis Ho: μ = 195g

3. Alternative hypothesis Ha: μ ≠ 195g ("≠" inequality because "effects of the new feed on the growth" is required to be analyzed)

4. Test statistic formula: t = (x-bar - μ)/(s/SQRT(n))

x-bar = estimate of the Population Mean (statistical mean of the sample) [178.62]

n = number of individuals in the sample [50]

s = sample standard deviation [14.84]

μ = Population Mean [195] (for computing Test statistic)

5. Computation of Test statistic from the formula t = (approx) -7.81

6. Determination of the P-value: The test is based on n -1 = 49 df (degrees of freedom). Table "look-up" value shows area under the 49 df curve to the left of t = -7.81 is (approximately) 0.0000

7. Conclusion: with significance value α = 0.01/2 (1 - 0.99 confidence)/2 the above shows P-value <= α, [0.0000 <= 0.01/2] so Null hypothesis Ho: μ = 195 should be rejected. Sample of 50 two-month old catfish with new feed formula shows to a 99% confidence the "true mean" weight is affected by new feed formula on two-month old catfish.  

With significance value α = 0.05/2 (1 - 0.95 confidence)/2 the above shows P-value <= α, [0.0000 <= 0.05/2].