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Test of proportion: 60 Minutes TV share

Short answer

Test H0: p=0.20 vs H1: p>0.20. Sample n=4860, p̂=1024/4860≈0.2107, SE=√(0.2×0.8/4860)≈0.00574, z=1.865. Critical value 1.645 at 5%; reject H0, evidence supports more than 20% share.

A simple random sample of Household with TV set in use. Show that 1024 of them were tuned to 60 minute while 3836 were tuned to some other show. Use 0.05 significant level to test the claim of CBS executive that "60 minute get more than a 20 share", which mean that more than 20% of set in use are tuned to 60 minute.

This is a right-tailed test of a single proportion (large sample).

n = 1024 + 3836 = 4860,  x = 1024
p̂ = 1024/4860 = 0.2107,  p₀ = 0.20, q₀ = 0.80
  1. H0: p = 0.20 (the share is 20%) H1: p > 0.20 (the CBS claim: more than 20% of sets are tuned to 60 Minutes)
  2. Level of significance: α = 0.05
  3. Test statistic:
SE = √(p₀q₀/n) = √(0.20×0.80/4860) = 0.00574
z  = (p̂ − p₀)/SE = (0.2107 − 0.20)/0.00574 = 1.865
  1. Critical value: z(0.05, right-tail) = 1.645. Reject H0 if z > 1.645.
  2. Decision: z = 1.865 > 1.645 (p-value = 0.031), so reject H0.

Conclusion: at the 5% level there is sufficient evidence to support the CBS executive's claim that 60 Minutes gets more than a 20% share.

✓ Answer checked by Claude Sonnet 5.5

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