<PRE><B>Question 867524</B>
 suppose a random sample of 1750 households with television showed that 69.5 percent of them have cable television. Is this sufficient evidence, at the 0.05 significance level, to claim more than two-thirds of all television households have cable televisions?
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Ho: p &lt;= 2/3
Ha: p &gt; 2/3 (claim)
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p-hat = 0.695
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z(0.695) = (0.695-(2/3))/sqrt[(2/3)(1/3)/1750] = 2.5143
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p-value = P(z &gt; 2.5143) = normalcdf(2.5143,100) = 0.006
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Conclusion: Since the p-value is less than 5%, reject Ho.
The test results support the claim.
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