Question

5. Information on a packet of seeds claims that 93% of them will germinate. Of the 200 seeds that were planted, only 180 germinated. (a). Construct a 95% confidence interval for the true proportion of seeds that germinate based on this sample. (b). Does this seem to provide evidence that the claim is wrong?

Answer 1

Given:

p0 = 93% = 0.93

n = 200

X = 180

The sample proportion is P = X = 180 - 0.9 n 200 T =0.9 Significance levelia =0.05 as 95% confidence interval is given by CI = + ZX12X SE Νοω The critical value of a at 95% confidence interval is 2X12 = 20.0512= 20.025 = 1.960 { SE = Poc!-fo) = 70.93(1-0.93) 200 = 0.01804 SE=0.01804] ..(J = Ř t2x127 SE = 0.9+1.960X0.01804 = 0.9 I 0.0354 CI=(0.8646, 0.9354)

Lower limit = 0.8646

Upper limit = 0.9354

b) Hypothesis test:

The null hypothesis, H0 : p = 0.93

The alternative hypothesis, Ha : p $\neq$ 0.93

Here, confidence interval includes 0.93, therefore we fail to reject null hypothesis, H​​​​​​0.

No, it does not seem to provide the evidence that the claim is wrong.